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ELEMENTS 



DIFFERENTIAL AND INTEGRAL 



CALCULUS 



WITH APPLICATIONS 



BY' 



WILLIAM S. HALL, E.M., C.E., M.S. 

PROFESSOR OF TECHNICAL MATHEMATICS IN LAFAYETTE COLLEGE 




<mnr. % 






,1' 



NEW YORK 
D. VAN NOSTRAND COMPANY 

1897 



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Copyright, 1897, 
By D. VAN NOSTRAND COMPANY. 



Xcrfoooti Press : 

J. S. Cushing & Co. — Berwick & Smith. 

Norwood, Mass., U.S.A. 



PEEFACE. 



This work is an introduction to the study of the Differential and 
Integral Calculus, and is intended for colleges and technical schools. 
The object has been to present the Calculus and some of its important 
applications simply and concisely, and yet to give as much as it is 
necessary to know in order to enter upon the study of those subjects 
which presume a knowledge of the Calculus. The book will be 
found to be adapted to the needs of the mathematical student, and 
also will enable the engineer to get that knowledge of the Calculus 
which is required by him in order to make practical applications of 
the subject. 

All of the formulas for differentiation are established by the 
method of limits. This method is preferred because it is more readily 
understood, and is more rigorous than the method of infinitesimals; 
and, moreover, it has the great advantage of being a familiar method, 
as the student has previously used it in Algebra and Geometry. 
But the differential notation is fully explained, and is employed 
when there is any advantage gained by so doing, particularly in the 
investigations of the Integral Calculus. 

As soon as the fundamental formulas of differentiation have been 
established, the corresponding inverse operations or integrations fol- 
low. Thus the essential unity of the two branches of the Calculus 
is emphasized, the whole subject is made more intelligible, and there 
is a saving of much space. 

Principal applications of the Calculus, as in Maxima and Minima, 



iv PREFACE. 

Radius of Curvature, etc., are treated at some length, while less 
important subjects are treated much more briefly. 

A large number of carefully selected examples, some original ones, 
and numerous practical numerical problems from mechanics and dif- 
ferent branches of applied mathematics are given. 

As there has been an increasing demand for a short course in 
Differential Equations, a chapter on this subject is given which it is 
hoped will meet a much-felt want. 

A table of Integrals, arranged for convenience of reference, is 
appended. 

Many American and English books, and some of the leading 
French and German works, have been freely consulted, and problems 
have been gathered from many different sources. 

WILLIAM S. HALL. 
Easton, Pa., January, 1897. 



CONTENTS. 



CHAPTER I. 

Definitions and First Principles. 

Art. Page 

1. Constants and Variables 1 

2. Functions 1 

3. Increments . 3 

4. Limits 4 

5. Theory of Limits 5 

6. Limiting Ratio of Increments 6 

7. Derivatives 8 

8. Differentiation and the Differential Calculus 10 



CHAPTER II. 
Differentiation of Algebraic Functions. 

9. Definition . 11 

10. Algebraic Sum of Functions 11 

11. Product of a Constant and a Function . . . . . .12 

12. Any Constant 12 

13. Product of Two Functions 13 

14. Product of Three or More Functions 13 

15. Quotient of Two Functions . . . 14 

16. Constant Power of a Function. Problems 15 

CHAPTER III. 
Differentiation of Transcendental Functions. 

17. Definitions 19 

18. Base of the Natural System of Logarithms 20 

19. Logarithmic Functions .20 

v 



vi CONTENTS. 

Art. Page 

20. Exponential Function with Constant Base . . • . . .21 

21. Exponential Function with Variable Base. Problems . . . .22 

22. Circular Measure . 24 

23. Limiting Value of — 25 

6 

24. Trigonometric Functions 26 

25. Inverse Trigonometric Functions. Problems ...... 29 



CHAPTER IV. 
Differentials. 

26. Definition ........ 34 

27. Geometric Interpretation of -^ . . -. . . . . .35 

dx 

28. Geometric Derivation of Formulas. Problems . . . . ,36 



CHAPTER V. 
Integration. 

29. Definition 40 

30. Fundamental Formulas . . . . . . . . . .40 

31. Elementary Rules 42 

32. Constant of Integration. Problems . . . . . ... 43 

33. Integration of Trigonometric Differentials. Problems .... 48 

34. Definite Integrals 49 

35. Geometric Illustration of Definite Integration 50 

36. Change of Limits. Problems . : . . . . . . . .52 



CHAPTER VI. 
Successive Differentiation and Integration. 

37. Successive Derivatives 55 

38. Successive Integration. Problems 56 

Applications in Mechanics. 

39. Velocity and Acceleration of Motion 59 

40. Uniformly Accelerated Motion. Problems . . . . . .60 

41. Derivatives of the Product of Two Functions. Problems ... 61 



CONTENTS. 



vn 



CHAPTER VII. 



Functions of Two or More Variables, 
the Independent 



A KT 

42. 

43. 
44. 
45. 

4G. 

47. 

48. 
49. 
50. 



Implicit Functions. 
Variable. 



Change of 



Partial Differentiation 

Total Differential of a Function of Two or More Independent Variables 
Total Derivative when u =f(x, y, z), y = 0(x), and z = 0i(x) 
Successive Partial Derivatives 

If u = fCx, ?/), to prove that -^- = -^L .... 

dydx dxdy 
Implicit Functions. Problems ...... 

Integration of Functions of Two or More Variables 
Integration of Total Differentials of the First Order. Problems 
Change of the Independent Variable. Problems 



Page 
64 

65 
66 



71 
72 
73 



CHAPTER VIII. 

Development of Functions. 

51. Definition 

52. Maclaurin's Tneorem. Problems 

53. Taylor's Theorem 

54. Demonstration of Taylor's Theorem. Problems 

55. Rigorous Proof of Taylor's Theorem 

56. Remainder in Taylor's and Maclaurin's Theorems 

57. Taylor's Theorem for Functions of Two or More Independent Variables 



78 
78 
83 
84 

87 



58. 
59. 

60. 

61. 
62. 



CHAPTER IX. 
Evaluation of Indeterminate Forms. 
Indeterminate Forms 



Functions that take the Form 



Functions that take the Form — 

CO 



Problems 



Functions that take the Forms Oxco and co — oo . 
Functions that take the Forms 0°, go and 1 ±0 °. Problems 
Compound Indeterminate Forms. Problems . 



91 

92 

94 

95 
96 



64. 
65. 



CHAPTER X. 

Maxima and Minima of Functions. 

Definitions and Geometric Illustration 
Method of Determining; Maxima and Minima . 



100 
101 



Viii CONTENTS. 

Art. Page 

66. Conditions for Maxima and Minima by Taylor's Theorem. Problems . 102 

67. Maxima and Minima of Functions of Two Independent Variables . . 112 

68. Maxima and Minima of Functions of Three Independent Variables. 

Problems 114 



CHAPTER XI. 
Tangents, Normals and Asymptotes. 



69. Equations of the Tangent and Normal 117 

70. Lengths of Tangent, Normal, Subtangent and Subnormal . . .118 

71. Tangent of the Angle between the Eadius Vector and the Tangent to a 

Plane Curve in Polar Coordinates 118 

72. Derivative of an Arc . - . . 119 

73. Derivative of an Arc in Polar Coordinates 120 

74. Lengths of Tangent, Normal, etc., in Polar Coordinates. Problems . 121 

75. Rectilinear Coordinates 123 

76. Asymptotes parallel to the Axis 124 

77. Asymptotes determined by Expansion . 125 

78. Asymptotes in Polar Coordinates. Problems . . . . . 125 



CHAPTER XII. 

Direction of Curvature. Points of Inflection. Radius of Curvature. 

Contact. 

79. Direction of Curvature 127 

80. Direction of Curvature in Polar Coordinates 129 

81. Points of Inflection. Problems 130 

82. Curvature. Problems ..'..-. 131 

83. Radius of Curvature 132 

84. Radius of Curvature in Polar Coordinates ...... 133 

85. Contact of Different Orders . . . . . . . .134 

86. Radius of Osculating Circle and Coordinates of Centre . . . .135 

87. Osculating Circle and Contact of the Third Order. Problems . . 136 



CHAPTER XIII. 

EVOLUTES AND INVOLUTES. ENVELOPES. 

88. Definition 139 

89. Equation of the Evolute ' ' . .139 

90. A Normal to Any Involute is Tangent to its Evolute . . . .141 



CONTENTS. ix 

Art. Page 

91. The Difference between any Two Radii of Curvature of an Involute . 142 

92. Mechanical Construction of an Involute 143 

93. Envelopes of Curves 143 

94. Equation of the Envelope of a Family of Curves. Problems . . 144 



CHAPTER XIV. 
Singular Points. 

95. Definitions . 148 

96. Multiple Points. Problems 148 

97. Cusps 151 

98. Conjugate Points. Stop Points. Shooting Points ..... 152 

CHAPTER XV. 
Integration of Rational Eractions. 

99. Rational Eractions. Problems ........ 155 

CHAPTER XVI. 
Integration or Irrational Functions. 

100. Irrational Functions 161 

101. Irrational Functions containing only Monomial Surds. Problems . 161 

102. Functions containing only Binomial Surds of the First Degree. Prob- 

lems 162 

103. Functions having the Form x2n+1(lx 163 

m 

(a + bx°~) 2 

104. Functions having the Form /(£, -T/^L+JM dx. Problems . . . 164 

V ^cx + dl 

105. Functions containing only Trinomial Surds of the Form Va + bx + ex 2 . 

Problems 165 

106. Binomial Differentials 167 

107. Conditions of Integrability of Binomial Differentials. Problems . . 167 

CHAPTER XVII. 
Integration by Parts and by Successive Reduction. 

108. Integration by Parts. Problems 171 

109. Formulas of Reduction. Problems 172 



CONTENTS. 



CHAPTER XVIII, 

Integration of Transcendental Functions. Integration by Series. 
Art. Page 

110. Introduction 178 



178 
179 
180 
183 



111. Integration of the Form \ f(x) (log x) n dx. Problems 

112. Integration of the Form I x m a nx dx. Problems 

113. Integration of the Form ( sin m 6> cos n dd. Problems 

114. Integration of the Forms ( x n cos(ax)dx and \ x n sin(ax)dx 

115. Integration of the Forms \ e ax sin n x dx and \ e ax cos n x dx. Problems . 183 

116. Integration of the Forms \f(jx) arc sin x dx, \f(x) arc cos x dx, etc. 

Problems 185 

117. Integration of the Form f — Problems . . . .185 

J a + b cos 6 

118. Integration by Series. Problems 187 

CHAPTER XIX. 
Integration as a Summation. Areas and Lengths of Plane Curves. 

119. Integration as a Summation. Problems . . . . . 188 

120. Areas of Plane Curves in Polar Coordinates. Problems . . . 192 

121. Rectification of Plane Curves referred to Rectangular Axes. Problems 193 

122. Rectification of Curves in Polar Coordinates. Problems . . . 196 

123. The Common Catenary 197 

CHAPTER XX. 
Surfaces and Volumes of Solids. 

124. Surfaces and Volumes of Solids of Revolution. Problems . . . 200 

125. Surfaces by Double Integration 203 

126. Volumes by Triple Integration. Problems 204 

CHAPTER XXL 

Centre of Mass. Moment of Inertia. Properties of Guldin. 

127. Definitions 207 

128. General Formulas for Centre of Mass. Problems . . . . 208 

129. Centre of Mass for Plane Surfaces. Problems . . . . .211 



CONTENTS. xi 

Art. Page 

130. Centre of Mass of Surfaces of Revolution. Problems .... 213 

131. Centre of Mass of Solids of Revolution. Problems .... 214 

132. Moments of Inertia of Surfaces. Problems ...... 215 

133. Guldin's Theorems. Problems ........ 216 



CHAPTER XXII. 
Differential Equations. 

134. Definition 

135. Differential Equations of the First Order and Degree. Problems . 

136. Homogeneous Differential Equations. Problems .... 

137. The Form (ax + by + c)dx + (a'x + b\y + c')dy = 0. Problems . 

138. The Linear Equation of the First Order. Problems 

139. Extension of the Linear Equation. Problems .... 

140. Exact Differential Equations. Problems ..... 

141. Factors Necessary to make Differential Equations Exact. Problems 

142. First Order and Degree with Three Variables. Problems 

143. First Order and Second Degree. Problems 

144. Differential Equations of the Second Order. Problems . 



218 
219 
220 
222 
223 
225 
226 
228 
231 
233 
235 



APPENDIX. 
Table of Integrals 239 



DIFFERENTIAL AND INTEGRAL CALCULUS. 

CHAPTER I. 
DEFINITIONS AND FIRST PRINCIPLES. 
Art. 1. Constants and Variables. 

The quantities employed in the Calculus belong to two classes, — 
constants and variables. 

A constant quantity is one which retains the same value through- 
out the same discussion. Constants are usually denoted by the first 
letters of the alphabet. 

A variable quantity is one which admits of an infinite number 
of values in the same discussion within limits determined by the 
nature of the problem. Variables are usually represented by the last 
letters of the alphabet. 

Art. 2. Functions. 

One variable quantity is a function of another when they are so 
related that for any assigned value of the latter there is a corre- 
sponding value of the former. Arbitrary values may be assigned to 
the second variable, which is then called the independent variable, 
while the first variable or function is called the dependent variable. 

For example, the area of a circle is a function of its diameter 
because the area depends on the length of the diameter, and the 
diameter whose length may be assigned at pleasure is the inde- 
pendent variable. 

b 1 



2 DIFFERENTIAL AND INTEGRAL CALCULUS. 

The trigonometric functions are functions of the angle, the angle 
being regarded as the independent variable. 

Expressions involving x, such as 

x 3 , ax 2 + bx -f c, log x, Vl — x 2 , 

are functions of the independent variable x. 

A quantity may be a function of two or more variables. For exam- 
ple, the area of a plane triangle is a function of its base and altitude ; 
the volume of a rectangular parallelopiped is a function of its three 
dimensions. 

The expressions, 



x^-Sx^f + y 3 , -Va 2 x 2 + b 2 y 2 , a x+y , 
are functions of x and y. 
The expressions, 

a 2 x 2 + bhf + c 2 z 2 , log (x 2 -f- xy + z 2 ), 

are functions of .r, ?/, and z. 

An explicit function is one whose value is expressed directly in 
terms of the independent variable and constants. For example, y is 
an explicit function of x in the equations 



y = - -\/x 2 — a 2 , and y = 2 ax + .t 2 + c 2 . 
Explicit functions are denoted by such symbols as the following : 

y =/(») ; 2/ = ^0) ; 2/ = 4> (*) ; y =/'(*) ; 

which may be read respectively: "2/ equals the / function of #"; "?/ 
equals the large F function of x " ; "y equals the cf> function of x " ; 
"2/ equals the /prime function of xP 

When the equation giving the relation connecting the variables is 
not solved with reference to y, y is an implicit function . of x. For 
example, y is an implicit function of x in the equations 

a 2 y 2 + b 2 x 2 = a 2 b 2 , ax 2 + bxy + cy 2 = 0. 



DEFINITIONS AND FIRST PRINCIPLES. 3 

Implicit functions are denoted by such symbols as the following : 
f(x,y) = V\ F(x,y) = 0; 4>(x,y) = ] 

which may be read, "the / function of x and y equals zero " ; etc. 

Art. 3. Increments. 

If a variable receives any addition to its value, this addition is 
called an increment, and is usually denoted by the symbol A placed 
before the variable. Thus an increment received by the variable x 
would be denoted by Ax, and would be read " delta x, ,} or " increment 
of x." The increment of a variable may be either positive or negative ; 
if it is positive the variable is increasing, and if it is negative the 
variable is decreasing. A negative increment is sometimes called a 
decrement. 

Art. 4. Limits. 

A limit of a variable is a constant value which the variable contin- 
ually approaches, and from which it can be made to differ by a quan- 
tity less than any assignable quantity, but which it cannot absolutely 
equal. 

For example, assume that a body is moving along a straight line 
from A to B as in Fig. 1, under the condition that in the first interval 

\ 1 2 3 4 B 

Fig. 1. 

of time it shall move one-half of the entire distance, or from A to 1, 
and one-half of the remaining distance, or from 1 to 2, in the second 
interval, and so on, moving during each interval one-half of the dis- 
tance remaining. In this case the entire distance AB is a constant 
toward which the distance traversed by the moving point continually 
approaches as a limit but never reaches. 

The limit of -,asa; increases indefinitely, is zero ; as a; in this frac- 
as 

tion increases the fraction decreases, and as x may be increased at 

pleasure, the fraction may be made to approach indefinitely near to 

zero. 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Let the locus of the equation y = - be drawn by the method of 

x 

rectangular coordinates as in Fig. 2. 
Y 




If x = 1, then y = 1 ; 
If x = 2, then y == .5 ; 
If # = 4, then ?/ = .25 ; 
If £ = 100, then y = .01 ; 



Or as the abscissa increases the ordinate decreases toward zero as 
a limit; thus the curve continually approaches the X-axis, but never 
reaches it. 

The limit of the value of the repeating decimal 0.555..., as the 
number of decimal places is continually increased, is f. 

A variable may approach its limit in three ways : 

1st. A variable may increase toward its limit, as is the case when 
a polygon is inscribed in a circle ; the polygon will increase toward the 
circle as its limit, as the number of sides is increased. 

2d. A variable may decrease toward its limit, as is the case when 
a polygon is circumscribed about a circle ; the polygon will decrease 
toward the circle as its limit, as the number of sides is increased. 

3d. A variable may approach its limit and be sometimes greater 
and sometimes less than its limit. For example, take the geometrical 



DEFINITIONS AND FIRST PRINCIPLES. 5 

progression whose first term is 1 and whose ratio is — J, giving the 
series 1, — -§-,-§-, — 2T> '" '■> nere * ne li m it °f * ne sum °f the series, as 
the number of terms is indefinitely increased, is f ; but the sum of 
any odd number of terms will be greater than this limit, and the sum 
of any even number of terms will be less. 

Art. 5. Theory of Limits. 

From the definition of a limit of a variable, it follows that the 
difference between the variable and its limit is a variable which has 
zero for its limit. Therefore, to prove that a given constant is the 
limit of a certain variable, it is sufficient to show that the difference 
between the variable and the constant has the limit zero. 

1st. The fundamental proposition in the theory of limits is the 
following : 

If two variables are equal and are so related that as they change they 
remain always equal to each other, and each approaches a limit, their limits 
are equal. 

Let x and y be the variables, and a and b their respective limits, 
and let x' and y' represent the differences between the variables and 
their limits. 

Then a = x-\-x\ and b = y + y'. 

Since x = y is always true, 

a - b = x* - y'. (1) 

In equation (1), x' — y' is equal to a constant, and x' and y' are varia- 
bles that approach zero as a limit. Hence x' — y' = 0, and, therefore, 
a — b = 0, or a = b. 

The supplementary propositions are readily established. 

2d. The limit of the algebraic sum of a finite number of variables 
is the algebraic sum of their limits. 

3d. The limit of the product of two or more variables is the product 
of their limits. 

4th. The limit of the quotient of two variables is the quotient of 
their limits. 



6 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Art. 6. Limiting Ratio of Increments. 

If an increment be given to x in y =f(x), y will receive a corre- 
sponding increment ; required the limiting value of the ratio — ^« 

Ax 
Taking first a particular function, for example, y = ax 2 . 

In this example, if x receives an increment represented by Ax or h, 

y will take a corresponding increment represented by Ay; and the 

equation becomes 

y + Ay = a(x + h) 2 

= ax 2 -f 2 axh + ah 2 . 

Subtracting y = ax 2 , 

Ay = 2 axh + ah 2 . (1) 

Dividing by Ax = h, 

*y = 2ax + ah. (2) 

Ax 

As h approaches zero, each member of this equation will approach 
a limit, and by Art. 5 these limits are equal ; therefore 

limit of ^ = 2ax. (3) 

Ax V J 

In order to make a definite application, let a = 16 in the given equa- 
tion, and substitute s for y, and t for x; then the equation becomes 
s = 16 1 2 , which is approximately the equation of a freely falling body 
near the earth's surface, s representing the number of feet fallen, and t 
the time of the fall in seconds. 

Then the proper substitutions made in equation (3) give 

limit of — = 32*, 
A£ 

which is seen to be the actual velocity at the end of t seconds. There- 
fore the limiting ratio of the increments of distance and time is the 
velocity at the end of the period. 

To illustrate further, let the object be to determine the increments 
produced in s by certain decreasing increments assigned to t, when t 
has some given value, as 10. 



DEFINITIONS AND FIRST PRINCIPLES. 7 

Substituting s = y, t = x=10, a = 16 and At = Ax, in (1), (2) 
and (3): 

As = 320 A* + 16 (A*) 2 , 

|? = 320 + 16 (A*), 

and limit of — = 320. 

At 

Let A* = 0.1, then As = 32.16 and — = 321.6 ; 

A* ' 

Let A* = 0.01, then As = 3.2016 and — = 320.16 : 

A* ' 

Let A* = 0.001, then As = .320016 and — = 320.016 ; 

At ' 

Let At = 0.0001, then As = .03200016 and — = 320.0016 ; 

A* ' 



And it is apparent that as At continually diminishes, As also 

As 

decreases, and the ratio of the increments, — , approaches 320 as its 

limit. 

Take next a geometrical example. 

Let the curve be the parabola whose equation is y — V2x, and 
whose locus is shown in Fig. 3. Let (V, y') be the coordinates of P, 
and (x' + Ax, y' + Ay) be the coordinates of any second point P'. 



Then y' + Ay = V2 (x' + Ax). (1) 

Subtracting y'=^/2~x', 



Ay = V2 (x' + Ax) - V2x' ; 



hence Ay = V2 (s' + As) - V2< (2) 

Ax Ax v ' 

Rationalizing the numerator of (2), 

Ay 2 Ax 2 



Ax Ax [V 2 (x' + Ax) + V2 x'] V2 (x' + Ax) + V2 



x' 



and limit of — ^ = ^^ = — == = — 

Ax 2V2x' V2x' V' 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



From the figure, it is obvious that — is the tangent of the angle 

Ax 

P'TN, and if the point P' approaches indefinitely near to P, the line 

P'T will be a tangent to the curve at P. Therefore, the limit of — ^, as 

Ax 




Fig. 3. 



Ax approaches zero, is the tangent of the angle which the curve makes 
with the X-axis, and is equal to the reciprocal of the ordinate of the 
point of contact. 

If P is at the extremity of the latus rectum, coordinates (-J-, 1), then 



limit of — ^ = 1 

Ax 



tan 45°, 



or the tangent to the parabola at the extremity of the latus rectum 
makes an angle of 45° with the X-axis, which is a well-known property 
of the curve. 



Art. 7. Derivatives. 

The limit of — ^ in the preceding article is called the derivative of y 

Ax 



with respect to x, and is denoted by 



dy 
dx 



Hence the definition : If y is 



a function of x, the derivative of y with respect to x is the limiting 



DEFINITIONS AND FIRST PRINCIPLES. 9 

value of the ratio of the increment of y to the corresponding increment 
of x, as the increment of x approaches zero. 

In general, let y=f(x). (1) 

When x is given an increment Ax or h, y takes a corresponding 
increment Ay, and the equation becomes 

y+Ay=f(x + h). (2) 

Subtracting (1) from (2), 

±y=f(x + h)-f(x). (3) 

Dividing (3) by Ax, *L = /fe±JWM. (4) 

Ax Ax w 

As Ax approaches zero, the limit of — ^ is the derivative of the 

dv Ax 

function, and is represented by -^« 

cix 

Therefore $L = 4M = limit of &±£=M. 
dx dx Ax 

The derivative is often called the differential coefficient, and the 

symbol f'(x) is frequently used instead of -^- 

cix 

The term "derivative" is fully as significant as differential coefficient, and is 
certainly to be preferred when the method of limits is used. The word "deriva- 
tive " will generally be used in this book. Derived function is another name which 
is sometimes adopted instead of the word "derivative." It must be carefully 
noticed that A and d are not factors, but symbols of operations. 

The general method of finding the derivative of any function of x is 
as follows : Two values of the independent variable, as x and x + Ax, 
are taken, and the corresponding values of the given function are 
found; the difference between these two values of y is the increment 
of the function corresponding to the increment Ax given to x. The 
limit of the ratio of these two increments, as Ax approaches zero, will 
be the derivative of the function. 

According to this method, general rules or formulas are obtained 
for forming the derivatives of the different kinds of functions. 



10 DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 8. Differentiation and the Differential Calculus.* 

The operation of finding the derivative of a function is called 
differentiation. 

The object of the Differential Calculus is to determine the deriva- 
tives of functions, and to investigate their properties and applications. 

* The Differential and Integral Calculus originated in the seventeenth century. 
Newton was the first discoverer of the new analysis, but to Leibnitz belongs the 
credit of priority of publication and the invention of a notation much superior to 
Newton's, and which has entirely superseded it. Leibnitz first published his new 
method in 1684. 

Newton called his method the method of fluxions. According to him, all quan- 
tities are supposed to be generated by continuous motion, as a line by a moving 
point. Fluxions are the relative rates with which functions and the variables on 
which they depend are increasing at any instant. 

Leibnitz considered all quantities to be made up of indefinitely small parts or 
infinitesimals ; a surface being composed of indefinitely small parallelograms, and 
a volume of indefinitely small parallelopipeds. The nomenclature and notation of 
the Calculus now in common use were original with Leibnitz, and introduced by 
him. 

According to Newton, the fluxion of x would be denoted by x, while by Leibnitz 

the corresponding derivative is — • 

dt 



CHAPTER II. 

DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 

Art. 9. Definition. 

An algebraic function is one in which the only indicated operations 
are addition, subtraction, multiplication, division, and involution and 
evolution with constant exponents. 

In this chapter only functions of a single independent variable x 
will be treated, and throughout the chapter u, v and w will be regarded 
as functions of x. 

Art. 10. Algebraic Sum of Any Number of Functions. 

If y be taken to represent the algebraic sum of three functions of 
x, the equation may be written 

y = u 4- v — w. (1) 

If an increment Ax is given to x, the variables y, u, v and w, which 
are functions of x, will take the corresponding increments Ay, Au, Av 
and Aw, respectively ; then (1) becomes 

y + Ay = (u -\- Au) -f- (v + Av) — (w + Aw). (2) 

Subtracting (1) from (2), 

Ay = Au + Av — Aw. (3) 

tv •-,. i A Aw Au . Av Aw ,.. 

Dividing by Ax, -9- = — + — • (4) 

Ax Ax Ax Ax 

When Ax approaches zero, 

limit of *£ = % limit of ^ = f ^, etc, by Art. 7. 
Ax dx Ax dx 

11 



12 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Therefore ^ = — + — - — , by Art. 5, 1st and 2d ; 

dx dx ax ax 

d(u -\-v — w) _du dv _dw T 

dx dx dx dx 

If the algebraic sum of four or more variables be given, the deriva- 
tive would be found similarly. 

I. Hence, the derivative of the algebraic sum of any number of func- 
tions of x is equal to the algebraic sum of their derivatives. 

Art. 11. Product of a Constant and a Function. 

Let a represent any constant, then the product of a constant and 
a function of x may be written 

y = av. (1) 

Let v and y take the increments Av and Ay corresponding to the 
increment Ax given to x, then 

y + Ay = a (v + Av). (2) 

Subtracting (1) from (2), Ay = aAv. 

Dividing by Ax; — ^ = a — • (3) 

6 J ' Ax Ax V J 

When Ax approaches zero, by Art. 5, 1st, 

limit ^ = limit (a *»\ 
Ax \ Ax) 

therefore -^ = a — , by Art. 7. II. 

dx dx 

If v = x, Av = Ax and -^ = a. 
dx 

II. Hence, the derivative of the product of a constant and a function 
of x is the product of the constant and the derivative of the variable. 

Art. 12. Any Constant. 

As the value of a constant remains unchanged in any one discus- 
sion, the constant receives no increment, or, in other words, the incre- 
ment of the constant is zero. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 13 

Let a represent any constant ; then 

Aa = and — = 0. 
Ax 

Therefore, when Ax approaches zero, 

— = 0. III. 

dx 

III. Hence, the derivative of a constant is zero. 

Art. 13. Product of Two Functions. 

Let the product of two functions of x be represented by y = uv. 
When x is given an increment, the variables v, u and y receive corre- 
sponding increments, and the equation becomes 

y + Ay = (u + Au) (v + Av) 

= uv + uAv -f vAu + Au Av. (1) 



Hence 


Ay= (v 


+ Av) Au 


+ uAv ; 




and 


** = (* 

Ax v 


1 A \ A™ 

Ax 


+ u Av - 
Ax 




When Ax approaches zero, 










limit ^ = ^ 3 

Ax dx 




limit u — = 

Ax 


dv 

--U-, 

dx 




limit (y + Av) = 


- v, and 


r .. Au 
limit — = 
Ax 


_ du 
dx 



(2) 
(3) 



Therefore, by Art. 5, 

dy d(uv) du . dv TTr 

-£ = — ^ — J -=v \-u — IV. 

dx dx dx dx 

IV. Hence, the derivative of the product of tivo functions of x is the 
sum of the products of each function by the derivative of the other. 

Art. 14. Product of Three or More Functions. 

Let the product of three functions of x be represented by y = uviv. 
The product of two of the functions, as uv, may be taken equal to 
z ; then, by the preceding article, 



14 DIFFERENTIAL AND INTEGRAL CALCULUS. 

d(uvw) d(zw) dz . dvj 
dx dx dx dx 

( du . dv\ , dw 

= w v — + u— + uv—~ 

\ dx dx) dx 

da . dv , dw T7 

= wv \-wu \-uv V. 

dx dx dx 

This process may be extended to the differentiation of the product 
of any number of functions. 

V. Hence, the derivative of the product of any number of functions 
of x is equal to the sum of the products of the derivative of each into 
the product of all the others. 

Art. 15. Quotient of Two Functions. 

u 
Let the quotient of two functions of x be represented by y = — 



Then 


vy = w, 


and, by V., 


dy , dv du 

v~ + y — = — 

dx dx dx 


Therefore 


du dv 
dy dx ' dx 
dx~ v 




du dv 

V u 

dx dx 



VI. 

V 

VI. Hence, the derivative of a fraction is equal to the denominator 
multiplied by the derivative of the numerator, minus the numerator multi- 
plied by the derivative of the denominator, divided by the square of the 
denominator. 

Cor. 1. If the numerator is constant, — = 0, by III., and VI. 

dx 

becomes 





dv 




u — 


dy 


dx 






dx 


V 2 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 15 

Hence, the derivative of a fraction with a constant numerator is 
negative and equal to the numerator multiplied by the derivative of 
the denominator, divided by the square of the denominator. 

Cor. 2. If the denominator is constant, — = 0, by III., and VI. 

dx 

becomes du clu 

dy _ dx _ dx 

dx v 2 v 

which is the same result that would be obtained by II. 

Art. 16. Constant Power of a Function. 

Case 1. When the exponent is a positive integer. 
Let v be a function of x, and n its exponent ; then 

y = v n , 
y + Ay = (v + Av) n , 
and Ay = (y + Av) n — v n . 

Expanding (v + Av) n , by the Binomial Theorem, and dividing by Ax, 

*y=[nv»-> + n ( n - 1 >v«-*(Av) ... +(A*)«>1— • 

Ax [_ 1-2 v ; -rv ; j Aa . 

When Ax approaches zero, Av approaches zero also ; hence 

&=**-•*. VII. 

dx dx 

Case 2. When the exponent is a positive fraction, — • 

n 

m 

Let y = v n ' 

then y n = v m . 

As m and n are positive integers, by Case 1, 

n yn i _y_ _ mv m i — . 
dx dx 

therefore ^ = m^dv = mv^_dv 

dx n y n ~ l dx n m -- dx 

n dx 



VII. 



16 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Case 3. When the exponent is negative and either integral or 
fractional, as — n. 

Let y = v~ n , 

then y = 



Differentiating by Art. 15, Cor. 1, 

n _idv 
nv — 

% = dx = _ m -n-ldV yj L 

dx v 2n dx 

VII. The derivative of a constant power of a function of x is equal to 
the product of the exponent, the function with its exponent diminished by 
unity, and the derivative of the function. 

Eadical expressions may be differentiated according to this rule, 
the quantities being first transformed into equivalent expressions with 
fractional exponents. 

The radical of the second order is the one that occurs most fre- 
quently. It is differentiated as follows : 

Let y — Vv = v 1 . 

dv 

dy . -idv dx 

dx dx 2Vv 

Hence, the derivative of the square root of a function of x is equal 
to the derivative of the function divided by twice the square root of 
the function. 

PROBLEMS. 

The formulas established in this chapter are sufficient for the 
differentiation of all algebraic functions of a single variable. 
Differentiate the following functions : 

1. y = a + bx + x s . 

dy d . 

_da, d(bx) d(x 3 ) , -r 
dx dx dx 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 17 

^ = 0, by III. ; ^) = b, by II. ; ^2 = Sx 2 , by VII. 
clx dx dx 

Therefore ^=6 + 3^. 
dx 

2. y=(a + x)(b + 2x>). 

( ^=(b + 2x*) *&±*± + (a + s) d ( 6 + 2a *> , by IV., 
c?x efce OX 

= (6 + 2 x 2 ) + 4 (a + x) x 
= 6 + 6 x 2 + 4 ax. 

40T 3 

3. v = • 

y (a + x 2 f 

Applying VI., and VII., 

dy = (q + x 2 ) 3 x!2r 9 -4f x3(a + a 2 ) 2 x (2s) 
da (a + a 2 ) 6 

= 12x 2 (a-x i ) 

(a + ar*) 4 

4. 2/ = x(l + ar 9 )(l+^ 3 ). 

S = (1 +x 2 )(l + a?>+*(l + a*)-f (1 + ar') + x(l +^)f (1 + or 5 ) 
rtic ckc aa; 

= (1 + af)(l + &0 + a?(l + af)(2 a?) + »(1 + ^)( 3a 

= l + 3ar + 4ar } + 6a; 5 . 

K w _ 1 + a; dy _ \-2x-x 2 

dx~ (l + x 2 ) 2 

dy _ a 

*M= (a + afp^(& + a^lX^ + a:) + n(a + a;)]. 
cZa; 

dy_ 1 . 

<ta (1 - a;) Vl - x 2 

dy__ n 
dx~~ x n+1 ' 

dy a + 3 x 

10. , = («-*) V« + *. ~ — ^=. 





^ l + x 2 


6. 


?/ = aVa\ 


7. 


2/ = (a+x) m (6 + x) n . 


8. 


*1 —X 


9. 


1 - 1 . 

y x n ' 



18 DIFFERENTIAL AND INTEGRAL CALCULUS. 

dy _ a 4 + a¥ — 4 x 4 



11. y = x (a 2 + x 2 ) Va 2 — x 2 . -r= . 

u K J dx ■y/a 2 -x 2 

x dy _ 1 



12. 2/ = 



x+^/l-x 2 dx Vl-^ 2 + 2x(l-x 2 ) 



13. y= VlH-s+Vl-s dy = 1+Vl-s 2 

14. ?/ = (1 - 3 x 2 + 6 a 4 )(l + x 2 ) 3 . ^ = 60 a?(l + x 2 ) 2 . 



15. ?/ = 



da? 
3 a 3 + 2 dy_ 



<x 3 +l) ¥ aa; o; 2 (x 3 + l)^ 



dx ~&i 



16. y = Vl+^+Vl-< ^ = _^fl + 



Vl+ar'-Vl-ar 2 dx a?\ VI- 

17. 2/ = 3(x 2 + l)^(4iC 2 -3). ^- = m^{x 2 + iy. 

CIX 

< lx 2 -\ dy l+4a 2 

18. y = • -£ = -• 

» Vl + a 2 ^ x 2 (l + a 2 ) 1 



,'.• 



i q *, _ V(^ + a) 8 ^2/ _ (x — 2a)Vx + a 

iy. 2/ — . ~r~ — g * 

Vo; — a ax (x — a) 11 

— 1 1 dy _ m(b + x) -f n(a + a;) 

^ ~" (a + a) m (6 + #) n * cte ~ (a + x) m +\b + a) n+1 ' 



21. 



_ / x \ n dy _ n 

V ~\1+VT^) ' dx~ x ^/± 



x- 



88 . y-J 1 -* . %>= 2 *( 2 -*> 



CHAPTER III. 

DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 

Art. 17. Definitions. 

All functions that are not algebraic are called transcendental. Tran- 
scendental functions are divided into four classes : 

1st. Logarithmic functions ; those in which a logarithm of a varia- 
ble is involved. 

2d. Exponential functions ; those in which a variable enters as an 
exponent. 

3d. Trigonometric functions ; those involving sines, cosines, tan- 
gents, etc., in which the arc is the independent variable. 

4th. Inverse trigonometric functions ; those derived from trigono- 
metric functions, by taking the arc as the dependent variable. Thus, 
from the trigonometric function, y = sin x, is obtained the inverse 
function, x = arc sin y, which is read, " x equals the arc whose sine is 
y." The inverse trigonometric functions are also called circular func- 
tions and anti-trigonometric functions. 

The inverse trigonometric functions are often expressed differently, 
as shown in the following identities : 

arc sin y = sin -1 y ; arc tan y = tan -1 y ; arc cosec y = cosec -1 y. 

This second notation, employed to express inverse trigonometric func- 
tions, was suggested by the use of negative exponents in algebra, but 
the student is cautioned against the error of regarding sin -1 ?/ as equiva- 
lent to Another application of this notation for inverse func- 

sm?/ 

tions is seen in an anti-logarithm; if y = logo;, then x = log -1 ?/. 

19 



20 DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 18. Base of the Natural System of Logarithms. 
The base of the natural, or Naperian, system of logarithms is the 
limit of ( 1 -j- - J as x approaches infinity. 



By the Binomial Theorem, 



3 



. ■ x \ xj\ X 

- 1 + 1 + T^+ 1.2-3 J + - 

When x increases indefinitely, 

limit (l + - Y= 1 + 1 + — + —1— + . . .. 
This limit is usually denoted by e. 
Therefore e = 1 + 1 + -L ■ + —±— + .... 

i • -i JL • Li • O 

By summing this series the value of e is found to be 2.7182818+, 
which is the base of the natural system of logarithms. 



Art. 19. Logarithmic Functions. 

Throughout this chapter, v and u will always be regarded as func- 
tions of a single independent variable x. 

Let the base of the system of logarithms be a ; 



let 


y = log a v; 


B 


y + Ay = log a (v + Av), 




Ay = log a (v + Av) — log a v 




= log. <« + *»> = lo Jl+ Av ) 

V \ V J 



V \ v J 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 21 

Av ± 

and ^ = ^log„fl+^V. 

Ax v V v J 

Now as Ace approaches zero, Av approaches zero; and therefore 

— approaches zero and — approaches infinity. 

If — be substituted for x in the preceding article, the limit of 

Av 

-I ] is seen to be e. 

v J 

dv 

Therefore ^ = log a e —. VIII. 

dx v 

Log a e is the modulus of the system in which the logarithm is 
taken and may be denoted by M. 

VIII. Hence, the derivative of the logarithm of a function of x is 
equal to the modulus, multiplied by the derivative of the function, divided 
by the function. 

Hereafter, when no base is specified, it will be understood that 
natural logarithms are used ; then 

M = log a e = log e e = 1, 

and VIII. becomes -? = VIII. a. 

dx dx v 



Art. 20. The Exponential Function with a Constant Base. 

Let the exponential function with a constant base be 

y = a\ 
Taking the logarithm of each member, 

log y = v log a. 
Differentiating by VIIL, 

dy 
^rdx , dv 

M J = losa Tx'' 



22 DIFFERENTIAL AND INTEGRAL CALCULUS. 

IX. 

10 

And when Naperian logarithms are used, 



therefore gg = a' logo dp. 

dx M dx 



dy ,. i dv TAr 

-£ = a" log a—- IX. a. 

da? da; 

If a = e in IX. a, since log e e = 1, 

dy v dv TV , 

-f- = e v —-> IX. 6. 

dx ax 

If v = a in IX. a, ^ = a* log a. (1) 

dx 

Ifa = ein(l), ^/ = e*. IX. c. 

dx 



IX. Hence, the derivative of an exponential function with a constant 
base is equal to the function multiplied by the logarithm of the base and 
by the derivative of the exponent, divided by the modulus. 



Art. 21. The Exponential Function with a Variable Base. 

Let the exponential function with a variable base be 

y = u v . 

Then log y = v log u ; 

dy du 

and by VIII., if— = M — - + log u —• 

J ' y u dx 

m , p dy ,,_i du , u v log u dv v 

Therefore -2 = vu v l 1 — P X. 

dx dx M dx 

X. Hence, the derivative of an exponential function with a variable 
base is equal to the sum of two derivatives ; the first being obtained as 
though the base were variable and the exponent constant, and the second 
as though the base were constant and the exponent variable. 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 23 

PROBLEMS. 



1. y = cc?logx. 

2. y = 



3. y = 

4. y = 

5. y = 

6. y = 

7. 2/ = 

8. y = 

9. 2/ = 
10. 2/ = 



og (2 # + a 3 ). 



a — x 






og^ + Vl + ar 2 ). 

Og# 2 . 
Og 2 X. 

og (log a;). 



'-"'g 



og 



Vic 2 + 1 - a 



Va,* 2 +1 + a; 



VI + ic+Vl — a; 



Vl + # — Vl — a; 



og(Vl + ^+Vl-^). 



11. y = 

12. 2/ = a e "'- 

13. y = a* 2 . 

14. ?/ = e x (x — 1). 

15. y = 2e^*(xi-3x + 6x^-6). 



^=z(21ogz + l). 
da; 



da; 


_2 + §a? 

2x + x* 


dy_ 


2a 


dx 


a 2 — x 2 


dy = 
dx 


1 
1-a; 2 


dy _ 


1 


dx 


■Vl+x 2 


dy = 
dx 


2 

X 


dy_ 


_ 2 log x 


dx 


X 


dy _ 


1 


dx 


x log X 


dy _ 


2 


dx 


vv + i 


dy _ 


1 


dx 


a; Vl — a^ 


dy _ 


.1 1 


dx 


% x VI - a; 4 


dy = 

dx 


= ofe log a. 


dy _ 
dx 


= 2 <x z " • log a • x, 


d& = 
dx 


- e x x. 


dy = 
dx 


-. xe V» 



24 DIFFERENTIAL AND INTEGRAL CALCULUS. 

e x — 1 dy 2 e x 



16. y = 

17. y = 



e + 1 

1 + a? 



dx 


(e* + l) 2 


dy _ 


xe x 


dx 


(1 + *) 2 


<%_ 


_ x x (1 — log a;) 



1 

18. 2/ = of. 

19 . y = x n a x . — = afx^ 1 (n + x log a). 

ClX 

20. ?/ = af. ^ = af(logx-|-l). 

ctx 

21. y = x }0 * x . ty- = logx 2 °x l °z x -\ 

dx 

22 . ?/ = af*. -^-z=x xX l log x + log 2 a? + - ]af . 

23. y = e x [a; n — naj n_1 +n(?i — l)a; n - 2 +-..]. c ^- = e x x n : 

Art. 22. Circular Measure.* 

In higher mathematics, angles are not measured by the ordinary 
degree or gradual system, but in terms of another unit. The circular 
measure of an arc of a circle is the ratio of the length of the arc to the 
length of its radius ; and it is evident that this ratio does not vary 
with the radius. Thus the value of an arc of 360° in circular measure 
is **L= 2tt, of 180° is tt, of 90° is | and of 1° is ~ 

The angle at the centre of a circle subtended by an arc equal to the 
radius is the radian or circular unit. 

Let x denote the number of degrees in an angle, and z the number 
of radians in the same angle; then since there are tt radians in two 

right angles, 

x _ z 
180 " tt' 

Therefore z = -|- x, 

180 

* Hall's Mensuration, §§ 9, 10, 11, and 12. 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 25 

A 180 

and x = z. 

■K 

Hence, to reduce from gradual to circular measure, the number of 

degrees in the angle is multiplied by -^- ; and to reduce from circular 

■ • • 180 
to gradual measure, the circular value is multiplied by 

7T 



Art. 23. Limiting Value of 



Let the small angle AOB in Fig. 4 be represented by 0, and the 
radius OA by a; and let BC, AB and AD be sin0, chord and tan0, 
respectively. 

D 




C A 
Fig. 4. 

The area of the triangle A OB = \o? sin0; 
The area of the sector AOB = ±a 2 6; 
The area of the triangle AOD = -|a 2 tan0; 

and these areas are obviously in an ascending order of magnitude ; 
hence tan > > sin 6, 

tanfl ^ 

or -^~n>-^—^> 1 ' 

sin 6 sin 6 

Thus - — - lies between -^— and 1 ; but when 6 approaches zero, 
sm0 sm0 ll 

- — - or approaches 1 ; hence, as diminishes indefinitely, — — - 

sin0 cos0 ..'./ sin0 

approaches the limit unity. 



26 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Art. 24. Trigonometric Functions. 

1. Differentiation of the sine of an arc. 

Let y = sin v, 

then y + Ay = sin (v + Av) ; 

therefore A?/ == sin (v + Av) — sin v. 

By Trigonometry, 

sin A - sinB= 2 cos %(A + B) sin J (.4 - B). 

Substituting v + Av = A, and v = B, in this formula, 

A o / . A?A . Av 
Ay = 2 coslv+— J sin — ; 

hence — = cos ( v -+- —-) 

Ax \ 2 J 



. Av 
sin — 

2 Av 



Av Ax 
2 

When Ax approaches zero, Av approaches zero, and by Art. 23, 

. Av 
sin — 

limit is unity. 

Av J 

2 

Therefore -^ = cos v — • XI. 

c&c die 

XI. Hence, the derivative of the sine of an arc is equal to the cosine 
of the arc multiplied by the derivative of the arc. 

2. Differentiation of the cosine of an arc. 

Let y = cos v, 

then Ay = cos (y + Av) — cos v. 

By Trigonometry, 

cos A — cos B = — 2 sin£ (J. + J5) sin %(A — B). 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 27 

Substituting v -f- Av = A, and v = B, 

■ Av 
sin — 

hence ^ = - sin A, + ^—1 *» 

Az V 2y Av Aa? 

2 

Therefore — = — sin v — XII. 

cfcc dx 

XII. Hence, £/*e derivative of the cosine of an arc is negative, and 
equal to the sine of the arc multiplied by the derivative of the arc. 

3. Differentiation of the tangent of an arc. 

sinv 



Let y = tan v = 



cos v 



d / • s d , x 

cost* — (sini') — smv — (cosv) 
E VI d_f sin v\ dx dx 



dx \cos vj cos- v 

cos 2 y f-sm-v— - 



cos- V 



= sec 2 v— • XIII. 

XIII. Hence, the derivative of the tangent of an arc is equal to the 
square of the secant of the arc multiplied by the derivative of the arc. 

4. Differentiation of the cotangent of an arc. 

T , cosv 
Let y = cotan v = —. 

sin v 

/ • dv\ f dv\ 

sin v [ — sm v — ) — cos v ( cos v — 

By VI d f cos v \ = V c I^l V dx J 

o dv VT^ 

= — cosec- v — XI \ . 

dx 

XIV. Hence, the derivative of the cotangent of an arc is negative, and 
equal to the square of the cosecant of the arc, multiplied by the derivative 
of the arc. 



28 DIFFERENTIAL AND INTEGRAL CALCULUS. 

5. Differentiation of the secant of an arc. 

Let w=secv=- 

cosv 

By Art. 15, Cor. 1, 

d , N dv 
— (cos V) Sill V 

dy _d L f 1 \ _ _ dx dx 

dx dx\cosvJ~ cos 2 i> ' cos 2 i> 

= sec v tan v — • XV. 

dx 

XV. Hence, the derivative of the secant of an arc is equal to the 
secant of the arc, multiplied by the tangent of the arc, into the derivative 
of the arc. 

6. Differentiation of the cosecant of an arc. 

Let y == cosec v = — 

smv 



dv 

cosv 

dy_d L f 1 V _ 



By Art. 15, Cor. 1, 



dx dx\smvj sin 2 i> 

= — cosec v cotan v — XVI. 

dx 

XVI. Hence, the derivative of the cosecant of an arc is negative, and 
equal to the cosecant of the arc, multiplied by the cotangent of the arc, into 
the derivative of the arc. 

7. Differentiation of the versed-sine of an arc. 
Let y = vers v = 1 — cos v. 

Then ^ = — (l-cosv) = sinv^. XVII. 

dx dx dx 

XVII. Hence, the derivative of the versed-sine of an arc is equal to 
the sine of the arc into the derivative of the arc. , 

8. Differentiation of the coversed-sine of an arc. 
Let y = covers v = 1 — sin v. 

Then ^ = -i(l--sihv)=-cosv— XVIII. 

dx dx dx 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 29 

XVIII. Hence, the derivative of the coversed-sine of an arc is nega- 
tive, and equal to the cosine of the arc into the derivative of the arc. 

Art. 25. Inverse Trigonometric Functions. 

It should be remembered that there are two ways of indicating 
the inverse trigonometric functions. The functions arc sin x, arc cos x, 
arc tan x, etc., are often written as follows: sin" 1 ^, cos -1 #, tan _1 x ? etc., 
respectively. 

1. Differentiation of y = arc sin v. 
Then v = sin y. 

By XL, ^ = cos2/^; 

dx dx 

hence dy = J_dv = 1 dp 

dx cos ydx Vl — sin 2 ydx 

dv 
Therefore <* (a*o sin*) = _gg_. XIX< 

dx Vl-v 2 

2. Differentiation of y = arc cos v. 
Then v = cos y. 

By XII., 

hence 



dv _ 
dx 


dy 
-smy-f; 
dx 




dy _ 


1 dv_ 
sin y dx 


1 dv 


dx 


Vl — cos 2 y^ x 




dv 




d (arc cos v) _ 


dx 




dx 


VI -v 2 





Therefore "v<— "; = _ . XX 



3. Differentiation of y = arc tan v. 
Then i? = tan y. 

By XIII., ^ = sec 2 / ?/ 



nence 



dx dx' 

dy _ 1 dv _ 1 cfa 

dec sec 2 ?/ cte 1 + tan 2 ?/ dx 



30 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Therefore 



d (arc tan v) _ 
dx 1 



dv 
dx 



V 



4. Differentiation of y = arc cot v. 
Then v =coty. 



By XIV., 

hence 



dv o dy 

— = — cosec y -*- ; 

dx dx 



1 dv 



1 dv 

dx cosec 2 ydx 1 -\- cotsm 2 y dx 



Therefore 



d (arc cot v) 
dx 



dv 
dx 



1 + v 2 



5. Differentiation of y = arc sec v. 
Then v = sec ?/. 



By XV., 

hence 

Therefore 



dv 



dv , dy 

— == secy tan y—\ 
dx dx 



dv 



dv 



dx sec y tan ?/ dx secy Vsec 2 y-ldx 

dv 

d (arc sec v) _ dx 
dx ~ v Vv 2 - 1 



6. Differentiation of ?/ = arc cosec v. 
Then v = cosec ?/. 



By XVI, 



dv dy 

— = — cosec y coty -£; 
do? da? 

dv 1 



dv 



hence -^ = 

da? cosec?/ cot ?/d# cosec?/ Vcosec 2 ?/ - 1 .*» 



Therefore 



d (arc cosec v) _ 
dx 



dv 
dx 



vVi 



XXL 



XXII. 



XXIII. 



XXIV. 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 31 

7. Differentiation of y = arc vers v. 
Then v = versy. 

VT , TT dv ■ dy i =- dy 

By X T II., — = sinv — = a 1 — cos-y-^; 

dx dx dx 

dy 1 dv 1 dv 

,;ence -^ = — — — = — ^===3- — 

dx -y \ _ co^ydx A 2 vers ?/ — vers 2 y - x 

dv 

Therefore tf (arc vers >■) = _^ => xxv _ 

*B A 2 I- - V 2 

8. Differentiation of # = arc covers v. 
Then « == covers y. 

Bv X\ III.. — = — cos v — = — vl — sm- y—: 
dx dx " dx ' 

, tfy 1 #H 1 dv 

hence -£ = - — = — 

c] - x a 1 — sin 2 2/ -- ; ' : v 2 covers ?/ — covers 2 ^/ r] - :c 

dv 

Therefore ^ (arc covers ,-) = J* ^yj 

dx \Z2v-v 2 



PROBLEMS. 

1 . u = sin nx. — = n cos h& 

2 . v = sin" a?. ^ = n sin n_1 a; cos x. 

dx 

3 . y = cot 2 (r s ) . — , = - 6 ^ cot (ft cosec2 ;3 ) c ^- 

4. v = los fsin 2 z). — = 2 cot;/:. 
J & v ; da; 

5 . v = sin 2 ;/: cos x. -^ = 2 cos 2 a; cos a; — sin 2 a; sin a?. 
9 dx 



6. >/ = e x cos ;/:. -^ = e z (cos a; — sin ;/:). 

J dx K J 



32 DIFFERENTIAL AND INTEGRAL CALCULUS. 

7. y = e cosx smx. ^-= e C03x (cosx - sin 2 x). 

dx 

8. y = smlogx. -£ = -cos (log a?). 

dx x 

9. y '== (cosa?) sLnx . -^ = (cosa?) sin *[cosa?logcosa? — since tan x]. 

eta? 

10. ?/ = log tana;. 



= lo § \f 



.-. + sin a; 
- sin » 

12. 2/ = iogseca?. 



13. i/ = arc sin-- 
a 



14. y = arc tan?- 

a 

1 — x 2 

15. v = arcsin 

J 1+a? 2 

16. ?/ = arc sin (3 a? — 4 a? 3 ). 

17. y — arc sec 2 a?. 

2x 



18. ?/ = arctan 



1-a? 2 

19. y = x Va 2 — a? 2 + a 2 arc sin 

20. i/ = e arctan *. 



21 . ?/ = Va 2 — x 2 + a arc sin -• 
J a 



rtrt 4. Vl — cos a? 

22. ?/ = arctan — « 

Vl + cos a; 



«#_ 


2 


cte 


sin 2 a? 


do? 


1 

cos a? 


da? 


= tan x. 


d?/_ 


1 


dx 


Va- 2 — x 2 


dy _ 


a 2 


dx 


a 2 + x 2 


dy _ 


2 


dx 


1+a? 2 


dy _ 


3 


dx 


VI -a? 2 


dy _ 


1 


dx 


a?V4a? 2 -l 


dy _ 


2 


da? 

cfa/_ 

da? 


1+a? 2 


= 2 Va 2 - a? 2 . 


dx 


p&rc tan a; 

1 + a? 2 


dy_ 
dx 


/a — a?V 
\a + xj 


dx 


1 

"2* 



DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. 33 



23. 2/ =arccot^ + log X El t . dy = 2az? 

x ^x -ha dx x 4, — q 



-h a dx x 4, — a 4 

x (arc sin a;) . , /= - 9 dii arc sin x 

24. ?/ = v y 4- log VI — or. -£.= 

Vl - x 2 *b ( X _ ^1 

n (^ + 1) 3 , 1 . 2a; -1 dy 

25. y = ±log K 3 , { +-— = arc tan — — -^ 

x 3 + 1 V3 V3 

' _ e ax (a sin x — cos a?) 

' V ~ a 2 -fl 



orr o ,j ^arc tan x 

28. ?/ = arcsec 



dx 


X s -hi 


dy _ 
dx 


= e ax sin x. 


dy_ 


garc tan x 



dx 1 -f- x 2 

x V5 cfa/ 1 

2 Va 2 + x - 1 *» .t Va: 2 + x - 1 



29. y = log \ —J— -f i arc tan a;. -^ = 

* 1 — x dx 1 — or 



CHAPTER IV. 

DIFFERENTIALS. 

Art. 26. Introduction. 

The formulas for differentiation given in the preceding chapters 
have been established by the method of limits. In this chapter another 
method of treatment will be presented which is called the method of 
infinitesimals. According to this second method, the independent varia- 
ble is supposed to change by the continued addition of an infinitely 
small constant increment. This increment is called the differential of 
the variable, and the corresponding increment of the function is called 
the differential of the function. The differential of a variable may 
then be defined as the difference between two consecutive values of the 

variable. Hitherto, the symbol -^ has been regarded as a whole, but 

dx 

here it is defined as the ratio of the differential of the function to the 
differential of the independent variable, and is regarded as a fraction. 
The phraseology and notation of the two methods are different, but 
they give identical results. To illustrate : 

Let y = x 5 , 

then by VII., ^ = 5x 4 . 

J ' dx 

If differentials are used, the equation becomes 

dy = 5 x 4 dx, 

which would be read, "The differential of y is equal to 5a 4 times the 
differential of x." 

In general, let -^ = f'(x), 

cix 

then dy = f'(%) dx. 

34 



DIFFERENTIALS. 



35 



Now the reason for sometimes calling the derivative the differential 
coefficient is apparent, as it is seen to be the coefficient of dx in the 
differential of f(x). 

If each member of each of the formulas, I.-XXVL, be multiplied 
by dx, a corresponding set of formulas will be obtained for the differ- 
entials of functions.* 



Art. 27. Geometric Interpretation of -^. 

dx 

The two methods may be compared geometrically. In Fig. 5, let 

AB represent any plane curve whose equation will show the relation 

between the coordinates of any point of the curve ; then the ordinate 

y may be expressed as a function of x, giving for the equation 

y = /(»)• 

1. By the method of limits. 

Let (x, y) be the coordinates of any point P of the curve and 
(x + Ax, y -J- Ay) the coordinates of any second point P f , and \j/ the 




Fig. 5. 



angle which the tangent to the curve at P makes with the X-axis. If 
be the angle which the secant through P and P' makes with the 
X-axis, and PM be drawn parallel to OX, 

* Lagrange in Mecanique Analytique says: "When we have properly con- 
ceived the spirit of the infinitesimal method, and are convinced of the exactness of 



36 DIFFERENTIAL AND INTEGRAL CALCULUS. 

PM Ax 

Now, suppose that P' approaches P, or, in other words, that Ax 
decreases toward zero ; evidently approaches \p, and tan approaches 
tarn// as its limit. 

By definition, limit — * = -^« 

Ax dx 

Therefore, by Art. 5, tan ty = -^« 

2. By the method of infinitesimals. 

Let P and P' represent consecutive points on the curve, then PN 
and P'N' are consecutive ordinates. The part of the curve, PP', called 
an element of the curve, is regarded as a straight line, and when pro- 
longed it forms the tangent to the curve at the point P. PM is drawn 
parallel to OX; 

then PM=NN'=dx, 

and MP' = dy. 

And as = if/, 

, , P'M dy 

tan \b = = -£. 

r PM dx 

Hence, the derivative of the ordinate at any point of a plane curve 
with respect to the abscissa is equal to the tangent of the angle which 
the tangent to the curve at that point makes with the X-axis.* 

Art. 28. Geometric Derivation of the Formulas for the 
Differentiation of the Trigonometric Functions. 

In Fig. 6, let AP represent a circular arc x, with radius = 1, and 
PP' = dx an infinitely small increment given to x. PS is drawn par- 
allel to OA, and PN and P'M are consecutive ordinates. 

the results by the geometrical method of prime or ultimate ratios, or by the ana- 
lytical method of derived functions, we may employ infinitely small quantities as a 
sure and valuable means of abridging and simplifying our demonstrations." 

* The student will be much benefited by plotting curves whose equations are 
of the form y =/(#), and interpreting the derivative obtained from each equation. 



DIFFERENTIALS. 



37 



PN = sin a; ; P'M— sin (x -f- da;) ; therefore P'JS = d sin a;. 
ON = cos a; ; OM = cos (x -f- dsc) ; therefore MN = — d cos jb. 
The triangle PP'S is a right triangle, and Z PP'S = Z POiV. 
Hence, cZ sin x = P'£ = PP' cos PP'£ = cos x dx, XI. 



and dcos# = 


- MN = - PP' sin PP'S 




= — sin x dx. XII 




u4T = tana;, 


and 


AT' = tan (a; + (to); 


hence 


TT' = cZtana;, 


and 


OT' = d sec a;. 




J3Z) = cot x } 


and 


BH = cot (x + da;) ; 


hence 


iZD = — d cot a;, 


and 


iTi? = — d cosec a;. 



From the triangles CTT' and iZD.E', 
similar to NOP, the differentials of the 
remaining trigonometric functions may be 
obtained. 

It will be noticed in this article, that 
the differential of a function is negative when the function decreases 
as the independent variable increases. 




PROBLEMS. 

1. If the side of an equilateral triangle increases uniformly at the 
rate of 2 inches per second, at what rate does the altitude increase ? 

Let x = a side of the triangle, and y its altitude ; then y 2 = J x 2 . 

/q 

Differentiating, and solving for dy, gives dy = —— dx, which shows that 
if an infinitely small increment is given to x, the corresponding incre- 
ment of y is 3—- times as great ; that is, the altitude increases ~~ times 

- 2 



38 DIFFERENTIAL AND INTEGRAL CALCULUS. 

as fast as the side. When x is increasing at the rate of ^— inches per 

second, y is increasing at the rate of —^~ times 2, or V 3 inches per 
second. 

Remark. In these examples the differentials are regarded as rates. 
The rate of change of a variable at a given instant may be here defined 
as the increment which it would receive in a unit of time, if its change 
should be uniform throughout the interval. Thus, when a variable at 
a given instant is said to change at the rate of 2 inches per second, the 
meaning is, that an increment of 2 inches would be added in one sec- 
ond, if the change should continue uniform for one second. 

2. If a circular plate of metal is expanded by heat, how rapidly 
does the area increase, when the radius is 2 inches long and increases 
at the rate of .01 inch per second ? 

Let x = radius, and y = area ; then y = 7r.se 2 , and cly = 2ttxcIx. 
When x = 2 inches, and dx = .01 inch per second, cly = .04 w square 
inches per second, which is the rate at which the area increases. 

3. The common logarithm of 1174 is 3.069668. What is the loga- 
rithm of 1174.8, if the logarithm is assumed to change uniformly with 
the number ? 

Let x — the number, and y = its logarithm ; 

then y = log x, and cly = — dx. 

x 

Hence, the increment of the logarithm is — times as great as the 

x 
increment of the number. 

Therefore dy = - 43429448 x .g = .000295. . . . 
And log 1174.8 = 3.069668 + .000295 = 3.069963. 

Remark. It will be seen from the equation dy = — dx, that as the 

x 

number increases by equal constant increments, the logarithm will 
increase more and more slowly. So the assumption made in the last 
example is not strictly true, but for comparatively small changes in the 



DIFFERENTIALS. 39 

number, the results are sufficiently accurate for practical applications. 
The use of the Tabular Differences in tables of logarithms is based on 
this assumption. 

4. In the parabola y 2 = 12x, find the point at which the ordinate 
and abscissa are increasing equally. Ans. The point (3.6). 

5. At what part of the quadrant does the arc increase twice as 
rapidly as its sine ? Ans. At 60°. 

6. The logarithmic sine of 30° 5' is 9.700062. What is the loga- 
rithmic sine of 30° 6' ? Ans. 9.700280. 

7. A boy is running on a horizontal plane directly towards the 
foot of a tower at the rate of 5 miles per hour. At what rate is he 
approaching the top when he is 60 feet from the base, the tower being 
80 feet high ? Ans. 3 miles per hour. 

8. A vessel is sailing northwest at the rate of 10 miles per hour. 
At what rate is she making north latitude ? 

Ans. 7.07 + miles per hour. 



CHAPTER V. 

INTEGRATION. 

Art. 29. Definition. 

Integration is the operation of finding the function from which a 
given differential has been obtained. The result of the integration is 
called the integral of the differential. The symbol which indicates the 
operation of integration is I . Since differentiation and integration 
are inverse operations, the symbols d and I , as signs of operations, 
neutralize each other.* 

The process of integration is of a tentative nature, depending on 
a previous knowledge of differentiation; just as division in arithme- 
tic is a tentative process depending on a previous knowledge of 
multiplication. 

For example, d (x 4 ) = 4 oc?dx ; 

therefore I 4jc 3 da; = x\ 



/• 



Art. 30. Fundamental Formulas. 

The fundamental formulas for integration are obtained directly 
from the formulas for differentiation. A function is the integral of 
a differential, if the function when differentiated produces the differ- 
ential. All integrations must ultimately be performed by the formulas 
of this article. When a differential is to be integrated, if it is not 
apparent on inspection what function when differentiated produces it, 

* The symbol J is derived from the initial of the word "summation." Leibnitz 
introduced the letter S to denote the operation, and this gradually became elon- 
gated into the symbol J". 

40 



INTEGRATION. 41 

the differential must be transformed into some equivalent expression 
of known form, whose integral is given by one of the fundamental 
formulas. 

1 . J (du + dv — dw) = u-\- v — w; from I. 

2. iadv = av, " II. 

3 . J nav n l dv = av n ; " VII. 

4. f^=alogv; " VIII. a. 
J v 

5. Ca v logadv = a v ; " IX. a. 

6. (e v dv = e v ; " IX. b. 

7. I cosvdv = sinv; " XI. 

8. I — sin v dv = cos v ; " XII. 

9. j sec 2 v dv = tan v ; " XIII. 

10. I — cosec 2 vdv = cotv; " XIV. 

11. I sec v tan v dv = sec v ; " XV. 

12. f — cosecycotvcfa; = cosecy; " XVI. 

13. J sin v dv = vers v ; " XVII. 

14. I — cos v dv = covers v ; " XVIII. 

15. f dv = arc sin v: " XIX. 

J a/1 _ o,2 



VI- v 

/ 



16. ^— = arc cos?;; " XX, 



Vl-v 2 



42 DIFFERENTIAL AND INTEGRAL CALCULUS. 

17. (- — — „ = arc tan v: from XXI. 
J 1 + v 2 

18. f_ *L_ = ar C cotv; " XXII. 

19. f — jg — = arc sec?;; « XXIII. 

•^ v ~>A> 2 — l 



c?v 

v -y/v 2 — 1 



20. C-— - 



21. f — -arc vers v; « XXV. 

J -\/2v-v 2 



dv 



22. f- 



Art. 31. Elementary Rules of Integration. 

The first four rules of integration will be demonstrated in full. 
(1) By L, d (u + v — w) = clu + dv — diu ; 

hence J d (w + 1? — w) = J (du + di> — dw), 

or u -\- v — w= I (du + dv — dw). 

But m.+ v — w = f du + I dv — I dw ; 

therefore I {du -\- dv — dw) = l du + I dv — I dw. 

Hence, the integral of the algebraic sum of any number of differen- 
tials is equal to the algebraic sum of their integrals. 



(2) By II., 


d(av) = adv ; 


hence 


| d (av) = J adv, 


or 


av= la dv. 


But 


av = a I dv 


therefore 


I adv = a I dv. 



INTEGRATION. 43 

Hence, the integral of the product of a constant and a differential 
is equal to the product of the constant and the integral of the 
differential. 

(3) By VII., dav n = nav n ~hlv. 
Then I dav n = J nav n ~^dv ; 

therefore I nav n ~hlv = av n . 

Hence, when a function consists of three factors, — viz. a constant 
factor, a variable factor with any constant exponent except — 1, and a 
differential factor which is the differential of the variable without its 
exponent, — its integral is the product of the constant factor, by the 
variable factor with its exponent increased by 1, divided by the new 
exponent. 

dv 

(4) By VIII. a, da log v = a — 

Then Cdalogv=C—; 

therefore J — — = a log v. 

Hence, the integral of a fraction whose numerator is the product 
of a constant by the differential of the denominator, is equal to the 
product of the constant by the Naperian logarithm of the denominator. 

Art. 32. Constant of Integration. 

By III., it is seen that the differential of a constant is zero ; hence, 
constant terms disappear in differentiation. Therefore, in returning 
from the differential to the integral, some constant must be added, 
which is called the constant of integration. The value of this arbitrary 
constant is determined in each case after integration by the data of the 
given problem, as will be shown hereafter. So, for the present, the 
undetermined constant will be omitted, but its addition after each inte- 
gration will always be understood. Frequently, when a differential is 
integrated by different methods, the results may not appear to agree, 



44 DIFFERENTIAL AND INTEGRAL CALCULUS. 

but on inspection it will always be found that the integrals differ only 
by some constant. 

PROBLEMS. 

Formulas 1-3. 

1. I ax?dx. 

By Formula 3, making v = x, and n = 4, 

I aaftfcc = \ I 4 ax 3 d# = 

2 . f 6 (6 ax 2 + 8 &a 3 ) 3 (2 aa + 4 6a 2 ) dx. 

d (6 ax 2 + 8 bx s ) = (12 aa + 24 bx 2 ) dx ; 

hence, if (2 aa? + 46a? 2 ) dx be multiplied by 6, it will be the differential 
of (6 asc 2 -f- 8 free 3 )" 3 " without the parenthesis exponent. After dividing 
the constant factor by 6 to preserve the same value, the integration 
may be effected by Formula 3, in which v = 6 ax 2 + 8 fr^ 3 . 

Therefore Cb (6 ax 2 + 8 bx 5 ) * (2 az + 4 5# 2 ) da 

= f - (6 ar 2 + 8 &r»)t (12 aa + 24 to 2 ) dx 

^(6a£ 2 + 8&a 3 )3 
= - g = ±(6ax 2 + 8bx*)i 

3 

3 r xdx = Ci( a 2 + X 2yh2xdx = (a 2 + a?jk 
J Va 2 + x 2 J 

4. I —\x~^dx. Ans. ^-x~^. 

5 . J (| ax% — f &#*) da;. Ans. ax^ — 6#i 
6 - J ~^=' u4ns. 2V#. 

7 - f(3S) dx - Ans - -l+A' 



INTEGRATION. 45 

Ans. 2( a 2 + aj 3)| 

C(6x* + 2 x 2 - 5) (3x 2 - l)dx. Ans. iM_lL^ + 5aj. 



r a 2 cfa 

^ (a 2 + ar 5 )* 



10. f(3 ax 2 + 4 fcjc 3 )* (2 ax + 4 6z 2 ) dec. Ans. i (3 ftz 2 + 4 6x 3 )i 
aV3to + 4cV 



n / * ft eta j 2 a V 3 fta; + 4 c 2 ce 2 

•/ W.S^ + 4-^ 2 3 bx 



12. 



| (6 — x 2 ) 3 x*dx. Ans. § & 3 £2 — £ & 2 a?2 + ^ &#Y" _ _?_ ^.^ 



Formulas 4-6. 



, 6 x 2 c?ic 



14 



■Si 

— - — Ans. log(x— a). 

x — a 

15 rx^dx ^ An ^ _l_ log / a + bx ny 

J a + bx n nb v y 

16 - f¥ri' Ans - log(^ 2 + f)i 

17. f(log^) 3 — • .4ns. J (logic) 4 . 

lg r5a?dx t Ans l0 g( 3a j4 + 7 )A # 
*/ 3 ar + 7 

19 -/Sl- A*.-J + f-Iog(.-l> 

20. fba 2 *dx = — — f ft 2 * • log a • 2 dx = -^— , by Formula 5. 
J 2 log a J 5 21oga J 

21. C3e x dx. Ans. 3e x . 

22. \ be ax dx. Ans. ~e ax . 
J a 



46 DIFFERENTIAL AND INTEGRAL CALCULUS. 

23. I 3 a x 'x log a • dx. Ans. fa* 2 . 

24. Ca x e x dx. Ans. — — — 
J 1 + log a 

Formulas 7-14. 



25. icosmxdx. 

/cos mx dx = — I cos mx • d mx = — sin ma;, by Formula 7. 
mj m J 

26. J sin 3 (2 a;) cos (2 a;) c?a;. 

I sin 3 (2 a?) cos (2x)dx=\ ( sin 3 (2x) cos (2a;) 2 da; 

= i f sin 3 (2 ») (2 sin (2 x) = | sin 4 (2 as). 

27. J sec 2 (ar 3 ) x 2 dx. Ans. -JtaiiaA 

28. 15 sec (3 x) tan (3 a;) dx. Ans. f sec (3 a;). 

/"sin (3 a;) da; A , 

29. I )±—J. Ans. isec3a5. 

J cos 2 (3 x) 

30. | e cosx sina;da;. Ans. —e C0Bx . 



31 



TgCOSX 

/(l + cos x) dx 
± — ■ : — L — • Ans. log I a; + sin x \. 

x + sin a; 

I tan x dx. Ans. log sec x. 



32. I tan a; da; 

33 . J sin 8 sec 2 6 dO. Ans. sec 

34. j cot x dx. 



. j cot a; da;. Ans. log sin a;. 

35. f*L = f £ = ri^!(M^ = iogtanix. 

J sin a; «/ 2 sin h x cos 4 a; J tan i a; 



f-*L = f_*2 = log tanfe + *« 

J cos* ^ S ing + x) V2 



37. 



INTEGRATION. 47 

I Ans. log tan x. 

J sin x cos x 



38. I . „ CX - — ^4?is. tana; — cot sc. 



■/, 



snr a; cos' a; 



39. r da? 

J Va 2 - b 2 x 2 



Formulas 14-22. 



dec & , 

-ax 
a 1 • bx 

= — arc sin — 



/ dx _ r a _i r 
Va?-b 2 x 2 ~J L b 2 x 2 ~ bJ L ¥x~ 2 b a 



L bh 



\ 



In order to integrate the preceding differential by Formula 15, it 
must be transformed into an equivalent differential having unity for 
the first term under the radical sign, and having for its numerator the 
differential of the square root of the second term under the radical 
sign. 

dx 



40. J — 

J x^Jb 2 x 2 — a 2 

— -dx 

/ dx _ r a _l r a 

xVb 2 x 2 -a 2 ~J lb 2 2 ~ a J b IW^ 
* a 2 a *a 2 



1 bx 

= - arc sec — , by Formula 19. 
a a 



41. f §2 Ans. - arc vers — . 

J V2abx~b 2 x 2 b a 

42. f 2xdx . A7is. arc sin (x 2 ). 

Jvr^aT 4 



44. 



Jdx i~~ 
' J.ns. arc cos (2 Va;). 
-\/X — 4:X 2 

C_xdx_ m Ans ^ i arcta n(a; 2 ). 

Jl + a; 4 2 v / 



45. j &x *dx __ t Ans 4V6arcvers(6^). 

^2x*- 6x% 



48 DIFFERENTIAL AND INTEGRAL CALCULUS. 

46. I — - — — - Ans. arc cot 



4 + x 1 2 

/dx 
- Ans. arc tan (x — 1). 
2 — 2 x -f- x' - 

AO C dx A 1 ex 

48. I - • Ans. — arc sec — 

J x -Vc 2 x 2 - a 2 b 2 ab ab 

AQ r dx 4 1 2 6 2 

49. I -• Ans. -arc covers — 

J -Va 2 x-b 2 x 2 & a 2 

50. f- dx 9 = f- ^-— = A arc tan (» + i) — 

Jl + x + x 2 Jf + (x + i) 2 V3 V3 



Art. 33. Integration of Trigonometric Differentials. 

Trigonometric differentials, to which the previous formulas cannot 
be made applicable by algebraic reductions, may often be brought 
to known forms by trigonometric reductions. 

Every function may be differentiated by a general method, but 
there is no general method of integration. Thus, while every function 
may be differentiated, but a limited number of differentials can be 
integrated. In attempting to integrate any given differential, the 
object is always to transform it to a fundamental form whose integral 
is known. 

Hence, the processes of the Integral Calculus are transformations 
to effect reductions to fundamental formulas. In order to become pro- 
ficient in these operations, it is necessary to have much practice in the 
solution of problems. 

PROBLEMS. 

1 . I cos 2 x dx. 

I cos 2 xdx = j (i + -A- cos 2 x) dx — \ x + \ sin 2 x. 

2. j sm*xdx. Ans. ^x — \sin2x. 

3 . J tan 3 x dx. 

j tan 3 x dx = I (sec 2 x — 1) tan x dx = \ tan 2 x — log sec x. 



INTEGRATION. 49 

4. I tan 2 x dx. Arts, tan# — x. 

5 . I tan 4 x dx. Ans. -J- tan 3 x — tan x -f- x. 

6 . I tan 5 x dx. Ans. \ tan 4 x — \ tan 2 x -f log sec #. 

7. I sin 5 xdx. 

isin 5 xdx = I (1 — cos 2 x) 2 since dec = I (— 1 + 2cos 2 £ — cos 4 x)d 



cos J x 



cos cc 

5, 



X. 



10 



8 . I cos 3 x dx. Ans. sin x — i sin 3 a;. 

9 . I cos 7 a; dx. Ans. sin as — sin 3 x + -f sin 5 x — \ sin 7 
. I cot 4 xdx. Ans. — -ir cot 3 x + cot £ + x. 

11. rcos 4 zcfo= f (i + icos2#) 2 cfa = i£ + isin2£-h-i- f cos 2 (2 a) d (2 a) 
= ^ # + i sin 2 ce + i [a; + Jsin4a;] = f a; + Jsin2a; + -^2-sin 4 ic. 



Art. 34. Definite Integrals. 

It was shown in Art. 32, that an arbitrary constant must be added 
after each integration. Before the value of this constant is determined, 
the integral is said to be indefinite. 

If, from the data of the given problem, the value of the integral is 
known for some particular value of the variable, the constant can be 
determined by substituting this value of the variable in the indefinite 
integral. 

For example, let 

ds 

— = at 4- v', in which a and v' are constants. 
dt y ' * 



50 DIFFERENTIAL AND INTEGRAL CALCULUS. 

By integrating and adding the constant C, 

S=±gf + v't+C. 
Now if S = S' when t = 0, C will be equal to S', 
and S = ±gt 2 + v't + S'. 

If, in any indefinite integral, two different values of the variable be 
substituted for the variable, and the result given by the second substi- 
tution be subtracted from the first result, the constant of integration is 
eliminated, and the integral is said to be taken between limits. 

The definite integral of f'(x)dx between the limits a and b is indi- 
cated thus : 

^ a f\x)dx, (1) 



jf 



in which a is the superior limit and b the inferior limit of integration. 
In (1), f'(x) dx is first to be integrated, then a and b are to be succes- 
sively substituted for x, and the second result is to be subtracted from 
the first. 

It is assumed that the integral is continuous between the limits a 
and b. A function is said to be continuous between two values of the 
variable when it has a single finite value for every value of the varia- 
ble between the given values, and changes gradually as the variable 
passes from the first value to the second. Evidently, the value of the 
integral up to the superior limit includes the value of the integral at 
the inferior limit. Hence, the difference between the values of the 
integral at two limits will be the value of the integral between those 
limits. 

Assuming that the inferior limit is equal to b, and writing the inte- 
gral in the two different ways ; 



fi 



f'(x)dx=f(x) + C, (2) 

and £f'(x)dx=f(x)-[f(x^ b (3) 

If these two forms are taken to represent the same quantity, 

fix) +C=f(x) - [/(*)],; whence C= [/(»)],. (4) 



INTEGRATION. 



51 



Thus it will be seen that the upper limit is any final value of the 
increasing variable x, and that the lower limit may be assigned without 
defining the upper limit. Equation (4) shows that the constant C de- 
pends on the lower limit. Therefore, the integral in equation (2) is 
indefinite because a free choice is left with regard to the selection of 
both limits. The part f(x) depends on the value of x selected for the 
superior limit, and the part C depends on the value taken for the infe- 
rior limit. 

Art. 35. Geometric Illustration of Definite Integration. 



The problem of finding the areas of plane curves was one of those 
that gave rise to the Integral Calculus, and this problem furnishes an 
illustration of the preceding article. 

In Fig. 7, let MN represent any plane curve ; it is required to find 
the area included between the curve, the X-axis and two ordinates. 

Let (x, y) be the coordinates of the point P. If BS = Ax be added 
to x, SQ = y + Ay. QC and PD are drawn parallel to the X-axis. 

If A represents the area of the curve between two ordinates and the 
X-axis, AA = area EPQS. 




Fig. 7 



Then 



PCQS = CR = y + Ay =1 Ay 
EPDS PR y y' 



Now, as Ax approaches zero, Ay also approaches zero 

and limit ^2^ = 1. 

RPDS 



y- Ax 


-^ 


dA _ 
ydx 


= 1; 


dA = 


= ydx, 



52 DIFFERENTIAL AND INTEGRAL CALCULUS. 

But the area RPQS is intermediate between area ECQS and area 

RPDS-, hence 

limit BPQS 

RPDS ' 

,■ •, &A 
or limit 

or 

therefore 

and A= Cydx+C. (1) 

If the area between the ordinates NW and ML is required, the 
superior limit will be the abscissa W, and the inferior limit will be 
the abscissa OL. If these limits are respectively a and b, the area will 
be denoted by 

A=£ydx. (2) 

Let the particular curve whose area is required be the common 
parabola, then y = ^/2px. Substituting this value of y in (1), gives 

A = CV2px dx+ C = V2p Cx^dx -f O — f V2px§ + C. 

If the area is estimated from the origin, when x = 0, A = 0; hence, 
by the first method of Art. 34, (7=0, and, therefore, A = ^-\/2px*. 

If the area is required between two ordinates whose abscissas are a 
and b, by the second method of Art. 34, 

j b ^/2pxdx = [% V2^] ft a = f V2p [a^ ;_&!]. 

Art. 36. Change of Limits. 

Let ff'(x)dx = f(x); 

then £ f\x)dx =/(a) -/(&) 

= -[/(&)-/<>)] 
= -£ f{x)dx. 



INTEGRATION. 53 

Hence, the limits of integration may be interchanged by changing the 
sign of the integral. 

It may also be readily shown that 

f f'(x) dx =C f (x) dx + f b f (x) dx. 

If a new variable be substituted for the old variable in integration 
between limits, corresponding changes must be made in the limits of 
integration. 

For example, I x n dx is required. 

Suppose x = z 2 ; then when x = 4, z = ± 2, 
and when x = l, z = ± 1. 



Therefore C x n dx = 2 f ^z 2n+1 dz 



4* +1 -l 

71 + 1 



PROBLEMS. 

1. Find the particular integral of dy=(x 3 —b 2 x)dx, if y=0 when x—2. 

Ans. y = t_^L + 2 b 2 -4. 

2 . Find the particular integral of du = (1 + J ax) *dx, if u = when # = 0. 

Ans. u = J- (1 + I az)' - -?-. 
27 V ^ 4 ; 27a 



2 4 



o ^ 



3. I (a 2 x — X s ) dx = 

I JU 

4. 2 f + ° (2-46)^ = 0.64. 8. f^ sin ^ = V2-l. 

»/0 Jo cos 2 

5. fWaa; = 38. 9. C^ldx = ±r. 
J2 Jo ^/ x 

6 . r^-„=f. io. r^ = =^. 

•/o a- 4- xr 2 a Jo yy _ ^ 

Jt 1+CB 2 2 Jo 12 



54 DIFFERENTIAL AND INTEGRAL CALCULUS. 



— ,. assume z = — ,4ns. 

Va 2 -ic 2 



a 



sin x cos 2 a? cto, assume sin x — z. Arts. \. 

14. In I — , assume y = l—x 2 . Ans. ± 1. 

Jo yi-a 2 

15. Find the area of the curve ?/ = x 2 + 6a;, between the abscissas 

x = 6 and aj = 0. ^4ns. 180. 

Note. Applications of the Integral Calculus in rectifying curves, determining 
areas, volumes, centre of mass, and moment of inertia, will be found in Chapters 
XIX., XX. and XXI. 



CHAPTER VI. 



SUCCESSIVE DIFFERENTIATION AND INTEGRATION. 



Art. 37. Successive Derivatives. 



As the derivative of a function is, in general, a new function of 
the independent variable, it can be differentiated. The derivative of 
the first derivative is called the second derivative. Likewise, when the 
second derivative is a function of the independent variable, it may 
also be differentiated, giving the third derivative; and so on. 



example, if y 


= ax A \ 


-£ = 4, ax 3 , 
dx 


dx \dxj 


d r d fdySP 
dx\_dx\dxj 


= 24 ax, etc. 



The symbols for the successive derivatives are usually abbreviated 
as follows : 

dx\dx) dxr 



dx 



' d fdy 
dx \dx 



dx\dx 2 dx? 



d fd n ~ l y \ = d n y 
dx\dx n - 1 ) ~ dx n 

The successive derivatives are often called successive differential 
coefficients. As the first derivative is often denoted by /'(#), the suc- 
cessive derivatives are often denoted by f"(x),f'"(x), etc. 

55 



56 DIFFERENTIAL AND INTEGRAL CALCULUS. 

If differentials are employed, successive differentials will follow 
instead of successive derivatives. The differential obtained immedi- 
ately from the given function is the first differential ; the differential 
of the first differential is the second differential ; and so on. If the 
function be represented by y, the successive differentials will be 
denoted by cly, dry, d 3 y, etc. 

In successive differentiation it is customary to make the assumption 
that the differential of the independent variable is constant ; i.e. the 
independent variable increases by equal increments, and hence is 
called an equicrescent variable. The independent variable will always 
be understood to be equicrescent unless the contrary is explicitly 
stated. 

Art. 38. Successive Integration. 

Two, three or n integrations must be performed in order to obtain 
the original function from which a second, third, or, in general, an nth 
derivative was derived. 

For example, if ^ = 24 ax. (1) 

Integrating (1), ^ = 12 ax 2 . (2) 

clx~ 

Integrating (2), ^ = 4 ax\ (3) 

ax 

Integrating (3), y = ax 4 . (4) 

But an arbitrary constant should be added after each integration, 
as a constant term may have disappeared at each differentiation. 

Then, in general, let — ^ =/(#); and denote the successive inte- 
grals of the function by f x (x), f 2 (x), f s (x), etc., and the constants of 
integration by C 1? C 2 , C 3 , etc. 

Given S =/{X) > 

then g* =/,(*> + <fc 



SUCCESSIVE DIFFERENTIATION AND INTEGRATION. 57 

dx n ~ 2 



&-M+M 



A@)+A=i+ c& + 'C» 



and finally, 



y =UX) + °' l- 2- 3 ••(»-!) + C2 1.2.3-( M -2) - + °" 



PROBLEMS. 

1. y =ax n . 2 - y =tana;. 

^ =na^" 1 ; ^ = sec 2 a; 






-— = u K a-± } ax , u^ 2 2 t 

c?z 2 ' 



^ = 6sec 4 a-4sec 2 z; 
dar 

— ^ = \n-a. — ^ = 8tanicsec 2 a;(3sec 2 cc— 1). 

dx n L dx 4 

3. y = ax 3 + foe 2 . 

4. y = log(a; + l). 

5. ?/ = a x . 

6. 7/ = 6a? 4 -4a 8 -5a?. 

1 

1 + ar 

8. ?/ = a? 4 log (a? 2 ). 

9. 7/ = tan 2 x -f 8 log cos x + 3 a 2 . 



dh, = 
dx 4 


= 0. 


d*y _ 
dx 4 


= _6(> + l)- 4 . 


dx n 


= a*(log a) n . 


d 4 y_ 

dx* 


= 144. 


d 3 y _ 


24a?(l-x 2 ) 


dx 3 


(1+a? 2 ) 4 


fy__ 
dx 5 


_48 # - 

X 


d 2 y _ 

dx 2 


= 6 tan 4 x. 



58 DIFFERENTIAL AND INTEGRAL CALCULUS. 

10. y — xe x . 



—z = (x -f n) e x . 
dx n K J 



11. y = e xc08a cos(xsma). -^ = e xcosa cos(# sina + a) ; 

(XX 

dx n v J 

12. g-a* y = g + ^ + C ^ +ft 

13. f| = 2x- 3 . 2/ = log«;+(7 1 | 2 +0 2a; +C3. 

dx s 2 

clii m n / y* 

14. —4 = sin a; cos 2 #. ?/ = f- C x sin a; -|- <7 2 . 

dec 2 6 

Let sin x= z: then — 4 = z. 



dz* 



15. ^ = C osz. 2/ = cos«+0 1 ^ + a 2 ^ + (73^+0 4 . 



. P + Yi _ ^2__£2^ r = o. 17. J* =.6369 a, when y = a sin oj. 

Ja ~ c V r ' Cdx 



d 2 x 
18. Find value of t from a — - = b (c — x). 

dt 2 

Integrating once gives 

a~ = b(2cx-x 2 )', 



Va 



Va 
i 



CX — X' 
X 



Therefore t = \j- arc vers 

b c 

19. In the harmonic curve whose equation is h = r x sinmZ + r 2 cos ml, 

d 2 h 
find — - ; r 1} r 2 and m being constants. -, 2h 

dl Ans. c ^ = -m 2 h. 

dl 2 



APPLICATIONS IN MECHANICS. 59 

APPLICATIONS IN MECHANICS. 
Art. 39. Velocity and Acceleration of Motion. 

The mean velocity of a moving body for a certain period, is equal to 
the distance passed over expressed in some unit of length, divided by 
the length of the period expressed in some unit of time. The velocity 
is uniform if equal distances are traversed in equal times ; and the 
velocity is variable if unequal distances are traversed in equal times. 

Let s = distance, v = velocity, and t = time. 

And let As denote the increment of distance passed over by the 
body in the increment of time At, while the velocity has increased to 
v -j- Av. 

The distance actually passed over, if the velocity is variable, lies 
between the distances it would have passed over if its velocities at the 
beginning and end of the period had been uniform ; hence 

vAt<As<(v + Av) At, 

As 

and v < — < (v -+- Av). 

Now, as At approaches zero, Av approaches zero, (v + Av) approaches 

v, and — approaches — ; and as the middle term is intermediate be- 
At dt 

tween the first and third terms, at the limit 

c ^ = v. (1) 

dt y } 

The acceleration at any instant is the rate at which the velocity is 
changing at that instant, and since the derivative of a function meas- 
ures the rate at which its value is changing, if the acceleration is 
denoted by a, 

dv /ON 

— = a; (2) 

dt ' w 

therefore a= *( d A = <^. (3) 

dt\dtj dt 2 v J 

Equations (1), (2) and (3) are fundamental formulas. 



60 DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 40. Uniformly Accelerated Motion. 

Motion is uniformly accelerated when the acceleration is constant. 
Denoting the acceleration, which may be positive or negative, by g ; 
then from (3), Art. 39, 

therefore — = gt + Q. (4) 



Cds = Cgt dt + C x dt ; 



therefore s = \ gt 2 + C Y t -j- C 2 . (5) 

If, in (4), zero be substituted for t, C x will be equal to I — ) ; or 

\dtJt=o 

G x will be the velocity at the beginning of the period, which may be 
represented by v . 

If, in (5), zero be substituted for t, C 2 will be equal to the left 
member ; or C 2 will be the distance already passed over at the begin- 
ning of the period, which may be denoted by s . 

Making these substitutions in (4) and (5), 

v = gt + v , (6) 

and s = i gt 2 + v t + s . (7) 



PROBLEMS. 

1. If a body is dropped, what distance will it fall in 5 seconds, and 
what will be its velocity at the end of the fifth second, the acceleration 
of gravity being 32.2 feet per second ? 

In (6) and (7), v = and s = ; hence 

v = gt, 
and s = | gt 2 . 

Substituting the values of g and t, gives 

v = 161 feet per second, and s = 402.5 feet. 

2. If a body is projected vertically upwards, to what height will it 
rise, and what will be the time of ascent ? 



APPLICATIONS IN MECHANICS. 61 

In this case, the acceleration is negative, and s = ; hence equa- 
tions (6) and (7) become 

v = -gt+v , (8) 

and s = — igt 2 + v Q t. (9) 

When the body attains its greatest altitude, its velocity becomes 
zero. Therefore, if v = in (8), 

t = v A (10) 

9 

which is the time during which the body rises. 

Substituting t = — in (9), gives 

which is the height to which the body will rise. 

3. A man is ascending in a balloon with a uniform velocity of 20 
feet per second, when he drops a stone which reaches the ground in 
4 seconds ; find the height of the balloon. Ans. 176 feet. 

4. A body is projected upwards with a velocity of 80 feet per 
second ; in what time will it return to the place of starting ? 

Ans. 5 seconds. 

5. Two balls are dropped from a balloon, one of them 3 seconds 
before the other ; how far will they be apart 5 seconds after the first 
one was dropped ? Ans. 336 feet. 

6. A body when first observed was falling at the rate of 40 feet per 
second, and struck the earth in 5 seconds ; required the entire distance 
that the body fell. 

Art. 41. Derivatives of the Product of Two Functions. 

Let y = uv, (1) 

u and v being functions of x\ then, by IV., 

dy du . dv /ON 

dx dx dx 



62 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Differentiating (2) with respect to x, gives 

dry _ dru, ch du du dv d 2 v 
dx 2 dx 2 dx dx dx dx dx 2 

d 2 u . dv du , d 2 v 

= v (-2 \-u 

dx 2 dx dx dx 2 

Similarly, 

d 3 y _ r d?u . dv dhi « d y d 2 u ~ d 2 v du d 2 v du d s v 
dx 3 dx 8 dx dx 2 dx dx 2 dx 2 dx dx 2 dx dx 3 

dhi . o dv dru . d 2 v du . d 3 v 

= v \-o J- o 1 u. 

dx 3 dx dx 2 dx 2 dx dx 8 



By proceeding as above, the fourth derivative, and other successive 
derivatives, may be obtained, and it will be seen that the same law of 
the terms applies, the numerical coefficients being those of the Bino- 
mial Theorem ; giving the general form, 

d n y _ d n u dv d n ~ 1 u n (n — 1) d 2 v d n ' 2 u 
dx n ~~ V dx^ n dxdx n ~ l 1-2 dx 2 dx n ~ 2 



d n ~ l v du d n v 
dx 71 ' 1 dx dx n 



. a- ~v uu . w"v /A x 

+ n — - u. (1) 



This is known as Leibnitz's Formula. 

That (1) is true for any nth derivative may be readily proved by mathematical 
induction. 

Differentiating (1), and arranging the terms, gives 

d n+1 y _ d n + l u . , ^ dv d n u n(n + 1) d 2 v d n ~hi 

dx n+1 ~ dx n+1 dxdx n 1-2 dx 2 dx' 1 - 1 

+ (n + 1) *5^ + ^ \u. (2) 

dx n dx dx n +^ 

If the law of the terms expressed by (1) is true for n, it then appears from (2) 
that the formula is true when n is changed into n + 1. But (1) has been shown to 
be true when n is 1, 2 or 3, then the formula must be true when n is 4, 5, 6, or any 
positive integer. 



APPLICATIONS IN MECHANICS. 63 



PROBLEMS. 

Find the derivatives in the following examples, by the aid of 
Leibnitz's Theorem : 

.31 d 4 y 1.2-3 

1. y = x 3 \ogx. — £ = 

LbtAs JU 

2. y = e ax z. -Jl = e «xl ct n z + na n l 1 — * ' a n 2 — - ■ 

dx 11 \ dx 1-2 dx- 

3. y = x 2 a x . ^ = a x (log a) n - 2 [(xlog a + ri) 2 -n~]. 



CHAPTER VII. 

FUNCTIONS OF TWO OR MORE VARIABLES. IMPLICIT FUNCTIONS. 
CHANGE OF THE INDEPENDENT VARIABLE. 

Art. 42. Partial Differentiation. 

If z be a function of two independent variables x and y, it may be 
expressed thus : 

*=/(*,»)• (1) 

In (1), z may be changed by changing either x or y. 
For example, in the equation of a plane, 

z = ax -f- by + c, (2) 

x and ?/ are two independent variables, of which z is a function. In 
(2) a value may be given to either coordinate x or y without any refer- 
ence to the other ; so if either x or y receives an increment, z will take 
a corresponding increment. Then z may be differentiated with respect 
to x and y separately. 

If (2) be differentiated, supposing x to vary and y to remain con- 
stant, the derivative is written 

dz /ON 

^= a - (3) 

If (2) be differentiated, supposing y to vary and x to remain con- 
stant, the derivative is written 

g=»- ^ 

These derivatives are called partial derivatives. 
According to the differential notation, equations (3) and (4) may be 
transformed into 

— dx = a dx, and — dy = b dy, 
dx dy 

and these expressions are called partial differentials. 

64 



FUXCTIOXS OF TWO OR MORE VARIABLES. 65 

Therefore, a partial differential of a function of several variables is 
a differential obtained on the hypothesis that only one of the variables 
changes. 

A total differential of a function of several variables is a differential 
obtained on the hypothesis that all of the variables change. 

To distinguish between the partial differentials of a function, the 

following notation is adopted : — Ax and — Ay will represent partial 

dz dz Ax . A ^ 

increments, and — dx and — dy partial differentials of z, with respect 
dx dy 

to x and y, respectively. 

D x z and f — ] have been used to represent total derivatives of z with respect 
\dx) 
to X. 

The general equation of a surface as given in Analytical Geometry 
of three dimensions is 

« =/(*> y), 

in which x and y are independent variables. 

If the surface be cut by a plane parallel to the XZ plane, the equa- 
tion of the curve of intersection will contain the variables x and z only, 

dz 
and the slope of the curve will be expressed by — 

dx 

Likewise, the section of the surface parallel to the YZ plane will 

dz 
contain the variables y and z only, and its slope will be — 

Art. 43. Total Differential of a Function of Two 
or More Independent Variables. 

Let z=f(x,y). 

Let x and y be given successive increments Ax and Ay, and repre- 
sent the corresponding total increment of the function by Az. 

Let z'=f(x-\-Ax,y)'j 

then £? Ax =f(x + Ax, y) -f(x, y), (1) 

Ax 

^- Ay =f(x + Ax, y + Ay)- f(x + Ax, y), (2) 

Ay 



66 DIFFERENTIAL AND INTEGRAL CALCULUS. 

and Az =f(x -f A#, y + Ay) - f(x, y). (3) 

Adding (1) and (2), and placing the first member of the resultant 
equation equal to the first member of (3), gives 

. Az A As' A //IN 

Az = — Ax H Aw. (4) 

Ax Ay . w 

Now, if Ax and A?/ approach zero, limit Az' = limit Az, 

therefore, dz = — dx -\ dy. 

dx dy 

Hence, the total differential of a function of two variables is equal 
to the sum of its partial differentials. 

Similarly, the total differential of a function of any number of 
independent variables may be found to be equal to the sum of its 
partial differentials. 

PROBLEMS. 

1 . z — ax s y 2 . 

-^dx = 3 a 2 x 2 y 2 dx ; -? cfo/ = 2 ax?dy. 
ox dy 

therefore dz = 3 axhfdx + 2 ax z y dy ; 

2 . z = x y . dz = yx y ~ x dx -f- x y log x dy. 

3. z = arc tan! dz = ^fllp . 

x x 1 + y 2 

4. z = sin (xy). dz = cos (xy) \_ydx + xdy~\. 

5. z = y 5 ' mx . dz = y 8inx logy cos xdx-\ sm x dy. 

y covers x 

6 . u = x yz . du = x yz ~ * (yz dx + zx log x dy -f- xy log x dz) . 



Art. 44. Total Derivative of u with respect to x when 
^ =/(»> V, z ), y=<f> (x), and z = ^(x). 

By Art. 43, 

7 du -, , du 7 . du 7 

du = — dx -\ dy -\ dz; 

dx dy dz 

, , o du _ du du dy du dz ^ , 

dx dx dy dx dz dx 



FUNCTIONS OF TWO OR MORE VARIABLES. 67 

Cor. 1. If u=f(x,y), and y = <£ (x), 

7 du 7 , du , 

du = — ax -\ dy: 

dx dy 

therefore du = 3u + ducly. 

dx ox ay dx ' 

Cok. 2. If u=f(y, z), y = <f>(x)y and z=<f> 1 (x), 

7 du , . di£ 7 
aw = — a?/ H dz ; 

therefore *• = *'& + *•* (3) 

dx 6?/ die dz dx 

Cor. 3. If u=f(y), and y=<fr(x), 

du = — dy; 
dy 

therefore d^dw#; (4) 

dec d?/ da? 

In the proposition, u is directly a function of x and also indirectly 
a function of x through y and z. 

In Cor. 1, u is directly a function of sc and indirectly a function of 
x through y. 

In Cor. 2, u is indirectly a function of x through y and z. 

In Cor. 3, w is indirectly a function of x through y. 



PROBLEMS. 

1. u = e ax (y — z), y = a since, and z = cos x. 

du_dududydu dz^ 
dx dx dy dx dz dx 

du ax/ N du ax die . 

— = ae ax (y —z), — = e ax , — = — e a 
dx KJ h dy dz 

dy dz 

-±=a cos x, — = — sin x ; 

dx dx 



68 DIFFERENTIAL AND INTEGRAL CALCULUS. 

therefore — = ae ax (y — z) -f ae ax cos x -f e ax sin x 

dx v ^ } 

= e ax (a 2 sin x — a cos # + a cos x + sin #) 
= e ax (a 2 -f 1) sin x. 

2. u = arctan (xy), and y = e x — - = _L_!±1_J. 

& 1 + x-e- x 

3. w = w«, 2/ = e x , and z=z 4 -4^-fl2a; 2 -24a;+24. — = e r z 4 . 

dx 

4. t£ = log (r 2 — y 2 ), and y = V sin ft — = — 2 tan ft 

do 

& ax (v z} t du 

5. m = — ^ s v = a sinx, and 2 = cosa?. — = e ox sm x. 

a 2 + l cto 



Art. 45. Successive Partial Derivatives of Two or More 

Variables. 

If u =f(x, y\ then — - and — - are, in general, functions of both x 
ox oy 

and y, and may be differentiated with respect to either independent 
variable, giving second partial derivatives. 

The partial derivative of — with respect to a? is — f — ) = — -• 

dx dx \dxj dx 2 

The partial derivative of — with respect to y is — ( — ] = — . 

By dy\dyj dy 2 

The partial derivative of -^ with respect to y is — I -^ ) = — — . 

ox oy\dxj dydx 

The partial derivative of — with respect to a? is — ( — ) = 

oy dx\dyj dxdy 

dhc 

Likewise, is a third partial derivative, obtained bv three suc- 

' difdx 1 ' J 

cessive differentiations ; first, with respect to x regarding y as constant, 
and then twice with respect to y regarding x as constant. 



a / d 2 u\ = d s u d_/ 



d ( d s u \ d A u 



dy\dxdyj dydxdy dx\dxdy 2 J dx 2 dy* 
and similarly with all other partial derivatives. 



IMPLICIT FUNCTIONS. 69 

ft 11 ft ii 

Art. 46. If u = f(x, y), to prove that = 

JK ' U)i dydx dxdy 

On the supposition that x alone changes in/(a;, y), 

Arc = f(x + Ax, y) -f(x, y) 

Ax Ax 



Elating (2) and (4), £(£)=£(£} 

Hence at the limits, — ( — ) = — f — V 
dy\dxj dx\dyj 

In the same manner it may be proved that 
d s u d 3 u d 3 u 



(i) 



Now, supposing y alone to change in (1), 

A (*¥\ = /Qg + Ao;, y + Ay)-f(x, y + Ay)-f(x + Ax, y)+/(a?, y) 
Ay\AuJ Ay - Ax 

On the supposition that y alone changes in/(«, y), 

Au _ f(x, y + Ay) -/(a, y) 
Ay Ay 

Now, supposing x alone to change in (3), 

A fAu\ f(x+Ax, y + Ay)—f(x+Ax, y)-f(x, y+Ay)+f(x, y) ^ 
Ax\AyJ Ax • Ay 



(2) 



(3) 



dx? dy dy dx 2 dx by dx 

This principle may be extended to any number of differentiations, 
and to functions of three or more variables. 



Art. 47. Implicit Functions. 

When in f(x, y) = 0, y can be expressed as an explicit function of x, 
the derivatives may be found by the methods already given. In this 
article a useful formula is established for obtaining the first derivative 
of an implicit function. 

Let u=f(x,y) = 0. (1) 

Then by Art. 44, Cor. 1, 

du _du du dy /o\ 

dx dx dy dx 



70 DIFFERENTIAL AND INTEGRAL CALCULUS. 

But u = 0, and therefore its total derivative equals ; hence 

|u + !»|/ =(X (3) 

dx dy ax 

du 
Solving (3) fori, gives | = -| (4) 

dy 
For example take x 2 + 2 yx + r 2 = 0. 
Then w = a; 2 -f 2 yx + r 2 . 

— = 2x + 2y, — = 2 x. 
dx . dy 

Therefore by (4), ^ = - ^±1^ = _ *±l m 

J w ' dx 2x x 

However, when an implicit relation between x and y is given from 
which y cannot readily be expressed as an explicit function of x, it is 
not necessary to resort to the method just given. But f(x, y) = may 

be immediately differentiated with respect to x, treating y as a function 

dy 
of x, giving what is called the first derived equation, from which ~ 

cix 

can be obtained as a function of x and y. For instance, given : 

x 3 — 3 axy -\- y 3 = 0. 
The first derived equation will be 

3x 2 -3ay-3ax ( ^ + 3y 2dy = 0. (1) 

J dx ' * dx w 

Solving (1) for ^, 
dx 



1 . u = cos (x + y) 

y 2 -f x 2 
2/^ — or 

3. u = arc tan f ^ » • 



%_ 


_ # 2 — a?/ 






da; 


aa; — 2/ 2 




PROBLEMS. 








verify 


3 2 u 
dydx 


d 2 u 
dxdy 




verify 


d 2 u 


dht . 




dxdy 


dydx 




verify 


dht 


d 3 u 




dy 1 dx 


dx dy' 1 



IMPLICIT FUNCTIONS. 71 

4. u = 6 e x ifz + 3 e y x*z 2 + 2 e *a?y - xyz , 



5% 



5. u == sin (a£ n -f 5?/ n ) ; verify 



d^ 2 dy dz 

d 4 w d 4 ^ 



12(e z 2/ + e y 2! + e*a;). 



d^ 2 d?/ 2 d?/ 2 3A3 2 



6. f - 2 ma*, +«•-« = a * = 5*=^ and ^{ = «("*'-!) . 

dec 2/ — m£C && (2/ — wise) 3 



7. y 3 — 3 y + x = 0. 



%_ 1 
d» — 3(1 — 2/ 2 )' 



8 . w + M m =1 . ^__^y*v-\ 



a) \bj dx \aj \y 



9. x 3 + 3axy + y 3 = 0. 



d 2 y _ 2 a 3 xy 
dx 2 (ax -\- y 2 ) 3 



Art. 48. Integration of Functions of Two or More Variables. 

Since integration is the inverse of differentiation, a partial deriva- 
tive is integrated by reversing the process of differentiation. 

d 2 u 
For example, the integral of — - =f(x, y) is found by integrating 

twice with respect to x, regarding y as constant ; but as y is regarded 
as a constant in this integration, it must be noticed that the constant 
of integration is an arbitrary function of y. 

Again, let it be required to integrate =/(#> y)- 

This may be expressed 

Evidently, in the second differentiation, — was differentiated with 

ox 



reference to y regarding x as constant ; therefore 

-ff(x,y)dy. (2) 



du 
dx 



In (2), u is evidently such a function that its derivative with 
•espect to x is j f(x, y) dy ; 



72 DIFFERENTIAL AND INTEGRAL CALCULUS, 

therefore, u = I I I / (x, y) dy \ dx, 

or u= I \ fix, y)dydx. 

In Art. 46, it was proved that the values of the partial derivatives 
are independent of the order in which the variables are supposed to 
change, hence the order of integration is also immaterial. 

d 3 u 



Similarly, if 


8y3xer f{X ' y) ' 


then 


u =ffCf(x, y) dydxdy; 


and if 


d 4 u ~, < 


then 


u = I J 1 1 f(x, y, z) dx 2 dy dz. 


r 


PROBLEMS. 


1. I I rdrdd-- 

2 


= 7 2i- 2 - S:£S> dxdydz=2 i 



Art. 49. Integration of Total Differentials of the First 

Order. 

If u be a function of x and y, by Art. 43, 

du = — dx-\ dy. (1) 

dx dy J W 

And from Art. 46, 4(g)=£g). (2) 

Therefore, if a total differential of a function of # and y is given of 

the form 

du = Pdx + Qdy, (3) 

then P=^, and Q=^, 

dx dy 

Hence from (2) ^=^> ( 4 ) 

dy dy 



CHANGE OF THE INDEPENDENT VARIABLE. 73 

which, is the condition that must be satisfied to make (3) an exact 
differential. This condition is called Euler's Criterion of Integrability. 
When (4) is satisfied, (3) is an exact differential of a function of x 
and y. Then the function u is obtained by integrating either term of 
(3); thus 

u^fPdx+fy). (5) 

In (5), the integration is with respect to x, hence the constant of 
integration is an arbitrary function of the variable which is treated as 

a constant, and f(y) must be determined so as to make — = Q. 

By 

For example, let du = 2 xy 3 dx + 3 x 2 y 2 dy. . 
Here P=2xf, and Q = 3x 2 y 2 . 

Hence, -— = 6 xy 2 , and — ^ = 6 xy 2 . 

' By y Bx J 

Therefore (4) is satisfied, and (5) gives 

u = j 2 xyhlx = aiy -\-f(y) = #V + c. 

PROBLEMS. 

1 . du = ydx + xdy. u = xy -\-c. 

2 . du = 4 x*]fdx + 3 x 4 y 2 dy. u = x^jf + c. 

3. du = J^ + {2y-^dy. u = ^ + y 2 + c. 

Art. 50. Change of the Independent Variable. 

Hitherto, the derivatives -^, — %, etc., have been obtained on the 

dx dx 2 

supposition that x was the independent variable and y the function, 
but it is sometimes advantageous to change the function into another 
one in which y is made the independent variable and x the function. 
And occasionally it is desirable to make a new variable, of which both 
x and y are functions, the independent variable. 



74 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



(a) To express — in terms of — 
7 dx dy 

If y is a function of x, then x may be regarded as a function of y, 
and ?/ may be treated as the independent variable. Evidently 

Ay Ax ^ 
-2 X — = 1; 

A# A?/ 

and as A?/ approaches zero, Ax approaches zero, and at the limit, 



therefore 



dy dx_ * 
dx dy 

dy = ]_ 

dx dx 

dy 



(1) 



d ii dx d x 

(b) To express -^ in terms of — - and — -, also to express 



terms of 



dx d 2 x 
dy dy 2 



dx 2 
and 



d 3 x 



dy dy* 



d 2 y_d_ fdy\ 
dx 2 dx \dx 



dx 3 



From (1), 

By Art. 44, Cor. 3, 



d 2 y 
dx 2 


d 

' dx 


1 

dx 
.dy \ 



d 
dx 


1 

dx 
dy . 


d 
~ dy 


1 

dx 
.dy\ 


dy 
dx 



therefore 



d 2 y 
dx 2 


d 
dy 


1 ' 
dx 
dy 


dy 
dx 


d 2 x 
dy 2 


dy 


d 2 x 
dy* 


fdx\ 2 


dx 


fdxV 


W 






W 



(2) 



CHANGE OF THE INDEPENDENT VARIABLE. 



75 



Similarly, —^ = — — - 



dx 2 ( 


d_ 
Ix 


" d 2 x 
df 

L(D'_ 


d 

dy 


d 2 x 
dtf 

fdx\ 3 


dy 
dx 




~fdx\ 3 d 3 x 3 fdx\ 2 fd 2 xY 
[dyj df [dyj [dtf) 


dy 




/dx\ Q 

\dy) 


dx 




fdx\d 3 x 3 fd 2 x\ 2 
K dy)dy 3 [dtf J 








fd£ 

\dy, 


r 







(3) 



In equations (1), (2) and (3), the independent variable is changed 
from x to y. 

(c) To express c -^-, c -^-, etc., in terms of c -^-, -\. etc., when x is 
dx dx- dz dz 2 

some given function of z. 
By Art. 44, Cor. 3, 



therefore 
and 



dy _ dy dz m 
dx dz dx' 

c]?y = ^fdy\ = ^/ r cty\dz. 
dx 2 dx V dx) dz V dx) dx ' 



(4) 
(5) 
(6) 



d 3 y _ d_ fd 2 y\ dz 
dx 3 dz \dx 2 J dx 

In equations (4), (5) and (6), the independent variable is changed 
from x to z. 



PROBLEMS. 
1. Change the independent variable from x to y in 

dx 2 \dxj dx 
Substituting the values of ^| and ^ from (1) and (2), 

(XX' (XX 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



drx 
fdx\ 3 



+ 



1_Y 
dx 

dy 



dx 
dy 



0; 



therefore 



#- 1+ g) 2 =°- 



2. Change the independent variable from x to z in 



en«= cos z. 






By (4), 


djf _ dy dz t 
dx dz dx 1 






dx , dz 
— = — sm z, hence — = - 
dz dx 


1 

sin z 


[ 


dy _ - 1 dy 

dx sin z dz 





(J) 



By (5), 



d 2 y _ d fdy\ dz m 
dx 2 dz \dxj dx ' 

d (dy\ d ( 1 dy s 



dz\dxj dz\ sin z dz J 






cos z dy _ 1 d 2 y 
sin 2 z dz sin z dz 2 



Therefore 



d 2 y _ _ /cos z dy _ 1 d 2 ?A 1 
d# 2 V sm2 2 dz sm z ^ 2 / si 11 z 



_ _ /'cos z dy _ 1 d 2 ?/ 
sin 3 z cfe sin 2 2 dz 2 



dy 



Substituting the values of -£ and — ^ from (7) and (8) in the given 



d 2 y 



(8) 



example, gives 



dx 



dx 2 



(l_COS^)fc% L 

\sin 3 z dz sin' 



z dz 2 



— cos 2 



(-1 

\ sm 



sin 2 dz 



Hence 



3-» 



CHANGE OF THE INDEPENDENT VARIABLE. 77 

3. Given y =f(x), and x = F(t), to express — , and —^ in terms of 

*?, *?, <&, and C ^. 
(ft d* 2 eft (ft 2 

By Art. 44, Cor. 3, dy = dyclx . 

J ' dt dx dt v ; 

dy 
dy dt 



Differentiating (1), and treating -£ as a function of £ through x, gives 

ax 

d 2 y _ d?y da? dy_ d 2 x 
dl?~dx 2 d? dxdi 2 ' 

d 2 y _ dy_ d 2 x d 2 y dx _ d 2 x dy 
d 2 y _ dt 2 dxdt 2 _df dt di?dt 
dx 2 ~~ dx?_ do^_ ^ ' 

o¥ d? 



CHAPTER VIII. 

DEVELOPMENT OF FUNCTIONS. 

Art. 51. Definition. 

A Function is said to be developed when it is transformed into an 
equivalent series. 

By the Binomial Theorem, constant powers of a binomial can be 
developed into series. For example, 

(1+x) i =1+ |_| + .|_... 

Some fractional functions may be developed by actual division. 
For example, 

— - — = l-{-3a + 9 x 2 + 27 #'.+ -.. 
1 — Sx 

The Calculus method of development is a general method, including 
the developments just given and many others as special cases. 

This is one of the most important applications of successive deriva- 
tives. 

Art. 52. Maclaurin's Theorem. 

Maclaurin's Theorem is a theorem by which a function of a single 
variable may be developed into a series of terms arranged according 
to the ascending integral powers of that variable, with constant co- 
efficients. 

The function to be developed is 

y =/(*)■ 

Assume the development of the form 

y =f(x) = A + Bx + Ox 2 + Dx s + Ex i -.. (1) 

78 



DEVELOPMENT OF FUNCTIONS. 79 

in which A, B, C, D, etc., are constants to be found by the method of 
Undetermined Coefficients. 

Forming the successive derivatives of (1) : 

^ =B + 20x + 3Da?+ 4Ex*+... (2) 

dx 

^L = 2C+2.3Dx+ 3-4J^ + '— (3) 

dxr 

^= 2-3J9 + 2.3-4 Bte+.» (4) 

die 3 



Since (1) and consequently (2), (3), etc., are assumed to be true for 
all values of x, they will be true when x= 0. Hence, making x = 
in each of these equations, and representing what y becomes on this 

hypothesis by (y) ; what -^ becomes by ( — ) ; what — ^ becomes by 
, 2 ,.\ dx \clxj dx 

— n ) : and so on : there follows 
dx 2 ) 

from (1), (y) = A, or A = (y) ; 

from (2), ( C ^]=B, or5=^ 



from( 3), (g)= 2 C, o,o = i i- 2 (g; 
^(4), (g)= 2 -?^« B -rb(S)' 



Substituting these values of ^4, 5, C, ••• in (1), gives 

^>=K1)H8)I + @)I + --- (5) 

If the function and its successive derivatives are expressed by 
fix), fix), f"(x), /'"(*), etc., 
equation (5) may be written, 



80 DIFFERENTIAL AND INTEGRAL CALCULUS. 

y=f(x) =/(0)+/'(0)|+/"(0)| 2 +/"'^+ "., (6) 

which is the formula of Maclaurin's Theorem.* 

If in the attempted development of a function by Maclaurin's 
Theorem, the function or some one of its derivatives becomes infinite 
when x = 0, the function cannot be developed by Maclaurin's Theorem. 
This is evident, because a finite function cannot be equal to a series 
containing infinite terms. 

PROBLEMS. 

1 . To develop y = (a + x) n . 
Here 

f(x) = (a + a) w ; nence /(0) = « n - 

/'(a;) =n(a + ic) ?l - 1 ; " /'(0) =m n - 1 . 

f"(x) = w(w - l)(a + x) n ~ 2 J " /"(0) = n (n - 1) a n ~\ 

f'"(x)=n(n - l)(n - 2)(a + ») n_3 ; " f"'(0)=n(n - l)(n - 2)a w ~ 3 . 

Substituting in (6), Art. 52, 

y =(a + a) n =a n -f m"-^ + '^r 1 ) fl^-V + w (^-l)(^-2) »-v+..., 

L? l£ 

which is the same development as that given by the Binomial Theorem. 

2. To develop y — log (1 + x). 

Here, f(x) = log (1 -f- a?) ; hence /(0) = 0. 

JL -j— X 

^ W (1 + a:) 2 ' ^ V ; 

/W(a . )= l_ilim « /'"(0)=1.2m. 

(1 + xf 



*This theorem is commonly known as Maclaurin's, having been first published 
by him in 1742 ; but as it had been given in 1717 by Stirling, it should more prop- 
erly bear the name of the latter. 



DEVELOPMENT OF FUNCTIONS. 81 

Substituting in (6), Art. 52, gives 

2 / = log(l + z)=™^-| + J-^ + 
and if the logarithm is in the Naperian System, 

log(l + »)= a: -| + -|-| , + -. 

Thus the logarithmic series is found to be a special case under 
Maclaurin's Theorem. 

3. To develop y = sin x. 

f(x) = sin x ; hence /(0) = 0. 

fix) =cosz; " /'(0) =1. 

/"(*)= -sins; " /"(0)=0. 

/'"(«) = -cos x; " /'"(0) = -l. 

Therefore, ?/ = sin a; = # — — -4- - — — -\ . 

' [3 [5 [7 

In the successive derivatives of sin x, the first four values are periodically 
repeating ; i.e., the fifth derivative equals the first, the sixth equals the second, etc.; 



hence, in general, 



(Z»(sin x) _ 
dx n 



(« + -|> 



Obtain by Maclaurin's Theorem the following developments : 

4. COS£ = l— ; . 

[2 li |6 



The general formula for the successive derivatives of cos x is 
— 5* L — cos \x + n-\- 

By the aid of the last two developments, natural sines and cosines 
may be computed. 

For example, to find the sine of 45°. 

By Art. 22, the circular measure of 45° is -• Substituting this 
value of x in the series of Prob. 3, gives 

4 [3.W 15 W [7.W 
= .7071068. 



82 DIFFERENTIAL AND INTEGRAL CALCULUS. 

X X Xp 

5. a x = 1 + log a - + log 2 a — + log 3 a — + .... 

1 If [£ 

If a = e and a? = 1 in this series, the value of the Naperian base 
may be computed. 

X Xi X 

6. arc tan a; = x ——+———+ •••• 

o o 7 

The labor in finding the successive derivatives may sometimes be 
lessened by expanding the first derivative by some one of the algebraic 
methods, as follows : 

f(x) =tan _1 cc. 

1 + x 2 
f"(x) = _ 2 # -J- 4 a 3 - 6 ar 5 + .... 

By substituting x = 1 in the development of arc tan x, 

4 3 5 7 

By Trigonometry, arc tan 1 = arc tan 1 -+- arc tan i. 



Hence - = 
4 



[l-I@' + l©'-'] + [l-l(i)* + Kl) , --3- 



Therefore tt = 3.141592 +. 

cc 2 x A , x 6 5x 8 



„. vi + ^ = l + |-| + ^-^ + 



/y.^ /y»t ™0 

8. e s[nx = l + x+~-— . 

2 2-4 3-5 

9. e x secic = 1 H-a? + a3 2 + -—H . 

O 

,. x* s , 1-3X 5 . 1.3.5a 7 , 
11. arc sin a? = oj + -—— + - — - — - + 



2-3 2-4-5 2.4.6-7 
It was by the use of this series that Sir Isaac Newton computed the value of ic. 



DEVELOPMENT OF FUNCTIONS. 83 

12. Develop y = logx. 

f(x) = log x-j hence /(0) = — oo. 
f'(x)= -; hence /'(0) = oo. 

/»(*) = - -3 ; hence /"(0)=- oo. 

Substituting in Maclaurin's Formula, gives, 

y = logx = — oo + go^ — oo — H . 

In this example, logic equals a series of terms involving oo, which 
makes the development indeterminate for all values of x. 

Hence this function cannot be developed by Maclaurin's Theorem. 

13. Develop y = cot x. 

14. If e be substituted for a in Ex. 5, 

Substituting xV-1 for x, 

= cos 03 + V — 1 sin x, by Exs. 4 and 3. (2) 

Substituting — x V— 1 for x in (1), gives similarly 

e -xV-i _ . cog # _ V — 1 sin #. (3) 

• Combining (2) and (3), gives 
since = 



,x\/-l _ e -x^-l 



and cos x = 



,2V- 1 
2 



These values of the sine and cosine are called their exponential 
values. 

The real functions, e% — e and ^_+A^, are called respectively the hyperbolic 
sine and hyperbolic cosine of x and are written sinh x and cosh x. 



84 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 53. Taylor's Theorem. 

Taylor's Theorem is a theorem for developing a function of the 
sum of two variables into a series of terms arranged according to the 
ascending powers of one of the variables, with coefficients that are 
functions of the other variable. 

Taylor's Theorem depends on the following principle which must 
first be established : The derivative of a function of the sum of x and 
y with reference to x regarding y as constant, is equal to the derivative 
of the function with reference to y regarding x as constant. 



Let 


u=f(x + y). 


Substituting 


z — x + y, gives 




«=/(*). 


In the first case, 


du _ df(z) dz 
dx dz dx 




=/'(?), since ^ = 1. 

OX 


In the second case, 


du _ df(z) dz 
dy dz dy 




=f'(z), since ~ = 1. 
dy 




du du 


Therefore 


dx = ty' 



This principle may readily be shown to apply in any particular 
example ; for instance, 
let u=(x + y) n , 



then 



du du / . n „_t 
t- = — = n(x + y) n \ 
dx dy 



Art. 54. Demonstration of Taylor's Theorem. 

Let u=f(x + y). 

Assume the development to have the form 

u = f (x + y) = A + By + Oif + Dif + . . . , 



(1) 



DEVELOPMENT OF FUNCTIONS. 85 

in which A, B, C, • • • are independent of y, but are functions of x. It 
is now required to find values of A, B, C, ••• by the method of un- 
determined coefficients. 

Differentiating (1), first with reference to x, regarding y as constant, 
then with reference to y, regarding x as constant, 

§¥ = ^A 4. dBy 4. ^tf + dBtf + ... 
dx dx dx dx dx 

^=B + 2Cy + 3Dtf+±Eif+.:. 
by 

O 7/ (ill 

But by Art. 53, — = — ; therefore, 
dx dy 

yL + My + M^ + dD tf = B + 2C 'y + ZDtf + ±Etf+.... (2) 
dx dx dx dx w 

Making y = in (1), gives A =f(x). 

Since (2) is true for every value of y; equating the coefficients of 
like powers of y in the two members by the principle of Undeter- 
mined Coefficients, 

^■=B, hence B = C V& =f (x) ; 

dx dx 

f x = 3D, hence Z> = ||(I/"(,)) = I/"'(, ); etc. 

Substituting these values of A, B, C, ••• in (1), gives 

u =f(x + y) =f(x) +f'(x) y +f'(x) y ~ +/'"(«) \{ + -, (3) 

which is Taylor's Theorem. 

If x= be substituted in (3), it reduces to 

m =/(0) +/'(0) y + f (0) t- +/»'(0) | 3 + -.., 



86 DIFFERENTIAL AND INTEGRAL CALCULUS. 

which is Maclaurin's Theorem. So Maclaurin's Theorem may be con- 
sidered as being but a special case of the more general one, Taylor's 
Theorem.* 

PROBLEMS. 

1. To develop (x + y) n . 

Substituting y = 0, and taking the successive derivatives, 
f(x) = x n , 
f'(x) =nx n ~\ 
f"(x) =n(n — l)x n ~ 2 , 
/"'(a?) = n(n-l)(n-2) x n ~\ 

Substituting these values in (3), 

(x + y y = *»-+ nx»-hj + n(7 j~ 1 ) >-y + n ( n -V( n ~ 2 ) x «-y + ..., 

which is the Binomial Formula. 

2. To develop sin (x-\-y). 

f(x) = sin x, f\ x ) = cos x, 
f"(x)=^sinx, f'"(x)= — GOSX. 

Therefore sm(x-\-y) = smxfl — r^ + ,-j ) + eosx(y — ^--\- ^ ) 

= sin x cos y + cos x sin y. 
Obtain the following developments by Taylor's Theorem : 

2 3 

3. a*+» = a x (l + log a • y + log 2 af-+ log 3 a^ + •••• 

4. log(aj + 2/) = loga! + ^-l^ + |^-.... 

x 2r 3 or 

5. (a + ^ = a;3 + Jaf%-iafV + A^'V • 

* Taylor's Theorem is named from its discoverer, Dr. Brook Taylor. It was 
first published in 1715, in a book by Dr. Taylor entitled Methodus Incrementorum 
Directa et Inversa. 



DEVELOPMENT OF FUNCTIONS. 87 



6. log(l + sin*) = z-|" + ?-^ + 



X 2 . X 3 X 4 

6~12 



7. log sec (x + y) = log sec x + tan x • y + sec 2 x - — 

+ sec 2 # • tana? • ^ — . 
3 



Art. 55. Rigorous Proof of Taylor's Theorem. 

In the demonstrations of Taylor's and Maclaurin's Theorems, it was 
assumed that the development would take place in a proposed form, 
and an infinite series was used without ascertaining that it was con- 
vergent. On account of these, as well as other objections, the method 
used is not altogether satisfactory. But, on the other hand, a rigorous 
investigation is necessarily complex and indirect. The proof which 
follows is one of the least difficult ones. 

The following proposition must be first established : 
If <f> (x) = 0, when x= a, and also when x = b, and if <£ (x) and 
4>'( x ) are nn ite and continuous between these values; then <j>'(x) will 
vanish for some value of x between a and b. 

The limit of —^--Jl, and hence -^ will have the same sign as — 
Ax dx dx Ax 

Ay 
when Ax is taken small enough. If y increases as x increases, — will 

be positive, and if y decreases as x increases, — will be negative. So, 

dy 
if -T- is always positive between the two given values of x, <£ (x) would 

be constantly increasing, and if -^ is always negative between the two 

dx 
values of x, <f> (x) would be constantly decreasing ; but neither suppo- 
sition can be true, as 4> (x) vanishes at the two given values for x. 

Therefore, <f>'(x) must change its sign between the two values, but 
a variable can only change its sign by passing through zero or infinity, 
and 4>'(x) remains finite by hypothesis; hence, <f>'(x) must pass through 
the value zero. 

, Let f(x) and its successive derivatives be finite and continuous 
between x = a, and x = a-\-h. 



88 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Assume 

* (x) =/(« + x) -f(a) - xf'(a) - g/»(a) ••• - .-/"(a) - ,-^iJ, (1) 

|SJ [n |n + l 

in which 

in + 1 



B = ^=[/(« + 7 -/(«) - A /'(") - 1/"(«) ■'" -|/"(«)]- 



(2) 



In (2), it is to be observed that R is independent of x. 
In (1), it is evident that <f> (x) = when x = 0, and when x = h. 
Hence, <f>'( x ) must be equal to zero for some small value of x between 
and h. Eepresent this value by x v 

Taking the derivative of (1) with respect to x, 

*'(*)=/'(« + x)-f'{a)-xf"(a)- %f»'(a) ... - f ^-f(a) -fs (3) 

= 0, when x = # 2 . 

But (3) also vanishes when x = ; hence there is some value of # 
between and Xj, for which <£"(o;) = 0. 

Continuing this process to n + 1 differentiations, 

^(a) =f«+\a + x)-R, 

for some value of x between and h, <j> n+ \x) = ; let this value of x 
be Oh, where 6 < 1, therefore 

/ n+1 (a + 0/0 = R. (4) 

Equating the values of R in (2) and (4), and solving for f(a + h), 

f(a + *)=/<?) + hf(a) + %f'{a)... + | i>(a) + -^-/»«(« + «). (5) 

[^ m [n + 1 

Now, since the only restriction imposed on a was that it must be 
finite, a may have any value ; hence x may be substituted for a in (5), 
which gives 

f(x + h)=f(x)+hf'(x) + 'lf"(x) ... +ff*(x) + ^-f'+\x+ 6h). (6) 

[IS [71 1 71 -f- 1 

From (6), Taylor's Theorem follows whenever the function is such 



DEVELOPMENT OF FUNCTIONS. 89 

that by sufficiently increasing n the last term can be made indefinitely 
small.* 

Art. 56. Kemainder in Taylor's and Maclaurin's Theorems. 

The last term of (6), Art. 55, p — -f n+1 (x + Oh), is called the re- 
mainder after n + 1 terms. 

For example, let f(x) = (1 + x) m , then by (6), Art. 55, 

(1 + X ) m = 1 + m + ffl (^- 1 ) ^ + ... 

+ — [m{m - 1) ... (m - n)(l + ftc)™-"- 1 ]. 

x n+i 
In this development, [m(m — 1) ••• (m — ?i)(l + 6>x) m_7l_1 ] is 

the remainder. ' 

If x is less than 1, the last term can be made indefinitely small by 
sufficiently increasing m. 

Hence, when x < 1, 1 + mx -\ ±— — ^# 2 +... is a convergent 

series. — 

If x = be substituted in the remainder in Taylor's Theorem, and 
then x be substituted for h, the remainder in Maclaurin's Theorem is 
obtained, which is 

If this remainder, when n is taken sufficiently large, becomes indefi- 
nitely small, Maclaurin's Theorem gives a convergent series. 

For example, let/(V) = sin a;. 



Then sin (x) = x- — + - - — H ^— sm 

[3 [6 [7 jn + 1 



te + (». + l)|T. 

But whatever may be the value of x, it is evident that 



sin 



|n + l 



ftB + (n + l)| 



* The proof of Taylor's Theorem given in this article is due to Mr. Hamershaw 
Cox. 



90 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



will have zero for its limit, hence this series is convergent for all values 
of x. 

Art. 57. Taylor's Theorem for Functions of Two or More 
Independent Variables. 



Let f(x, y) be a function of two independent variables, and suppose 
f(x + h, y -f k) is to be expanded in ascending powers of h and k. 

Regarding y as constant, and expanding as though x was the only 
variable, 
f(x + h,y + k)=f(x, y+k)+hj-f(x, y+k) + ^^f(x,y+7c) + .-.. (1) 

Expanding fix, y -\-k), regarding x as constant, and y as the only 
variable, 

f(x,y + k)=f(x, y) + k^-f(x, y) +~ 2 f(x, 2/) + -. (2) 

Substituting this value off(x, y + k) from (2) in (1), gives 
fix + h,y + k) = f(x, y) + h ^/(», V) + k yf(x, y) 



4- 



),£_ f(X) y)+mJL y M ,)+*£/<, y)] 



ff> 



*wM + - 



Similarly, 
fix + h,y + k,z + l) 



fix, y, z) + h —f(x, y,z)+k —f(x, y,z)+l £/(x, y, z) 
1 

12 
i 



/i2 £ /(ajj y ' 2 ) +2M ^^^ 2 ) +fc2 §^2/^)+---] 



P-^/fe 2/, «) + 3 » -r^-fix, y, z) + 
dard?/ 



dx 3 ' 



+ •••■ 



And in like manner a function of any number of independent vari- 
ables may be expanded. 



CHAPTER IX. 

EVALUATION OF INDETERMINATE FORMS. 

Art. 58. Indeterminate Forms. 

A function of x is indeterminate when the substitution of a par- 
ticular value for x gives rise to one of the following expressions : 

GO AO ..O-l+oo 

-, — , oo — 00, , 00 , 1 . 
oo 

The true value of a function which becomes indeterminate is the 
value which the function approaches as its limit, as the independent 
variable approaches the particular value which makes the function 
indeterminate. 

For example, to find the true value of — — — when x = a. 

x — a 

When x — a, this fraction assumes the form -• 
If a + h is substituted for x, the fraction becomes 

(«+ft)'-a' = 2a _k 

a + h — a 

Now if h approaches zero, the independent variable approaches the 
particular value a, and the function evidently approaches 2 a as its 
true value. 

/v»2 q2 

Again, if both numerator and denominator of the fraction 

x — a 

are divided by x — a, the quotient is x -f a, and now when x = a, the 
true value is found as before to be 2 a. 
As another example, 



* = ?, whenz = 0. 



x 2 O 7 

91 



92 DIFFERENTIAL AND INTEGRAL CALCULUS. 

By rationalizing the numerator, 



a — Va 2 — x 2 _ a 2 — (a 2 — x 2 ) 
x 2 ~ x 2 (a+Va 2 -x 2 ) 



a -f- Va- — x 2 _ 



2 a 



By algebraic and trigonometric transformations the true values of 
many indeterminate forms can be readily found, but the Differential 
Calculus furnishes a method applicable to all cases. 

Art. 59. Functions that take the Form -• 

Let f(x) and <f>(x) be two functions, such that f(x) = and <£(x) = 0, 
when x = a ; 

then m = ± 

<f>(a) 

Let x take an increment h ; then by Taylor's Theorem, 



f{x) + f'( x )h+f"(x) ! £+f"'(x)'£ + ... 
f(x + h) = [2 E 

*■(» + *) <jX» + *!(*)» + <#>"(*) ! l + 4>'"(x) ! £+... 

l£ IS 



(1) 



Substituting a for aj, making /(a) = and <f> (a) = 0, and dividing 
both terms of the fraction by h, 

h ^7^ f'^+f" ( ^h+f'"(fl)%+- 

/((> + h) = E l£ , 9 , 

Hence, as /i approaches zero, by Art. 5, 

/(q) = /'(q) 

0(a) </>'(a) 

which is the true value of Vt-t when a; = a. 

</>(V) 

If /'(a)=0, and <£'(a)=0, then ZM = 9, and the result is still 

^> (a) 

indeterminate. In this case, dropping the first term of the numerator 



(3) 



EVALUATION OF INDETERMINATE FORMS. 93 

and also of the denominator of (2), dividing both terms of the fraction 



by — , as h approaches zero, 



/(«) /"(«) m 

4(a) ~ 4,>/ l ' 



If (4) is also indeterminate, the process is repeated until a ratio of 
two derivatives is obtained, both of which do not reduce to zero when 
x = a. 

If f H (a) = and <£"(«) be not 0, the true value is 0. 

If f n (a) be not and <£ n (a) = 0, the true value is oo. 

PROBLEMS. 

X 5 — 1 

1. Find the true value of — , when x = l. 

x — 1 

f(x)=v>-l, </>(»)= x — 1; 

hence, fX x ) = 5 4 > and (f>'(x) = 1 ; 

therefore 4^ = ~^ = ^ = 6, when x = 1. 

2. Find the true value of 





■\ a? ' 






f(x) _ x — sin x 
<f>(x) x z 


_0 

0' 


when cc = ; 


f'(x) _ 1 — cos X 
<f>'(x) Sx 2 



~0' 


when x = ; 


f"( x ) _sinx 
<f>"(x) 6x 



~0' 


when cc = ; 


f'"(x) _ COS X 
<t>'"(x) 6 


1 

~6' 


when x = ; 


x — sin x~ 


1 






X 3 


x = 6 







therefore 



The subscript denotes the value which is to be substituted for x in the function 
within the brackets. 



94 DIFFERENTIAL AND INTEGRAL CALCULUS. 



3. 



log a; 
, ct' — b x 

4 - -IT' 

e x _ e -x 

5. : , 

since 

1 — x n 



6. 



7. 



1-x' 

tana; + seca; — 1 
tan a; — secaj-f 1 ? 

e x — e~ x — 2 x 

8. ; , 

x — sin a? 

x 3 — ax 2 — a 2 a? + o? 
ar — or 

10. ax-x 2 ^ 

a 4 — 2 a 3 x + 2 aa; 3 — sc 4 

tana? — sin a? 



11 



when x = 1. 


^.ns. 1. 


when a; = 0. 


Ans. log- 

6 


when # = 0. 


-4w*. 2. 


when x = 1. 


-4ns. n. 


when a; = 0. 


^4ns. 1. 


when a; = 0. 


Ans. 2. 


when aj = a. 


Ans. 0. 


when a; = a. 


^4ns. — oo. 


when a; = 0. 


Ans. ^. 



surx 

Art. 60. Functions that take the Form — . 

oo 

When x = a, & function may take any one of the forms 

oo a "oo 

_, _, or — . 

a oo oo 

Evidently, _ = oo, — = 0, and — is indeterminate. 

a oo oo 

Let ZM = ^, when a; = a. 

<f>(x) 00 

By taking the reciprocals of f(x) and <f> (as), 

1 



/to = £to = W henx = a. 
<l>(x) _J_ 0' 

/to 



EVALUATION OF INDETERMINATE FORMS. 95 



Therefore, by Art. 59, 



df 1 \ $'(*)_ 



f(x) 4(x) dx\<i>(x)j [$(x)y = <i>'(x)Lf(x)y 

+ (x) j_ ±{_i_\ fM /'(*) [*(*)]* 
/(*) dx\f(x)J [f(x)y 

and when x = a, /W = *'(") [/(«)]' (1) 

hence ,/W = /W (2 ) 

<£(«) *'(a) 

Therefore, the true value of the indeterminate form — can be found 

' GO 

by the same method as that of the form -. 
J 

However, in dividing (1) by Zi^i, it was assumed that 1S®1 is not equal to 
0(a) 0(a) 

or co . But (2) gives the true value in these cases also, as may be shown as 
follows : ., . 

Suppose the true value of A^V t0 be °> and let ^ be a finite quantity ; then 
iF 0(a) 

/(q) + h = /(a) + fe0(a) = ft 

0(a) 0(a) 

jj u t JK a )+ — 0W k as a va i ue which is neither nor go , hence (2) will apply 
0(a) 
to it, giving 

f(a)+h</>(a) f'(a)+h<t>'(a) 



0(a) 0'(a) 

0(a) 0'(a) 



therefore 



/(a) = /(a) 

0(a) 0'(a) 



Similarly, if the true value of ; { =co when x = a, then ^-— = 0, and the 
. .- ,- 0(a) /(a) 

same demonstration applies. 

Art. 61. Functions that take the Forms x oo and co — oc. 
Let f(x) x <j> (x) = x co, when x = a. 

Then mxHa) =m° ; 



<f>(a) 
therefore, the true value may be found as in Art. 59. 



96 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Again, let f(x) — <j> (x) = oo — go, when x = a. 

The expression in this case can be transformed into a fraction, 

which will assume either the form - or — , and the true value is found 

°° 
as before. 

For example, to find the value of 

sec x — tan x, when x = ^- 

, 1 — since i 7r 

sec x — tan x = = -, when x = — 

cos a; 2 



Therefore 



r/Ml 



— cosce 

— sin x 



= 0. 

2 



Art. 62. Functions that take the Forms 0°, oo°, and 1 ±0 °. 

Let f(x) and <j> (x) be two functions of x, which, when x = a, take 
such values that [/(#)]** is one of the assumed forms. 

Let y=[>]* x ; 

then log y = <£ (x) \ogf(x). (1) 

1st. When f(x) = oo or 0, and <f>(x) = 0, (1) becomes 

*(a)log/(aj) = 0(±co). 

2d. When f(x) = 1, and </> (a;) = ± co , (1) becomes 

+ (*)16g/(*) = (±oo)x0. 

Therefore, the true values of the logarithms of all the functions 
which take the forms, 0°, c©°, and 1 ±0 °, may be obtained as in Art. 61. 
For example, to find the value of x x when x = 0. 

Let y = af ; 

then log y = x log x = (— oo ), when x = 0. 

Hence, log y = °%f = — — . when x = 0. 

x -i CO 

-f(logz) 1 

LAW Ju i- -i r\ 

therefore log x x = 0, when a? = ; hence x x = 1, when a; = 0. 



EVALUATION OF INDETERMINATE FORMS. 97 

PROBLEMS. 

Find the true values of the following functions : 



1. 


logcc 




when x = ao . 


2. 


tana; 
3*' 




when &• = Jt. 


3. 


1 — log X 

e x 




when a? = 0. 


4. 


log tan 2 a; 
log tan x 




when x = 0. 


5. 


x n log X, 




when a? = 0. 


6. 


2*sin-^, 




when a; = oo. 


7. 


secxf a; sin x — 


i) 


when a; = -• 

2 


8. 


2 1 

x 2 - 1 x-i' 




when a; = 1. 


9. 


x 1 




when x — 1. 


log x log X 


10. 


cosec 2 a; -, 

X 2 




when a; = 0. 


11. 


2 1 




when a; = 0. 




sin 2 x 1 — cos a/ 


12. 


©■" 




when a; = 0. 


13. 


X.S1BZ 




when x = 0. 


14. 


1 
(^ + I)", 




when a; = oo . 


15. 


H)' 




when a; = oo. 


16. 


sina; tani: , 




when # = 7r - 

2 



Ans. 


0. 


Ans. 


3. 


Ans. 


0. 


Ans. 


1. 


Ans. 


0. 


Ans. 


a. 


Ans. 


-1. 


Ans. 


_ i 

2' 


Ans. 


1. 


Ans. 


1 
3- 


Ans. 


1 

"2- 


Ans. 


1. 


Ans. 


1. 


Ans. 


e. 


Ans. 


e a . 



Ans. 1. 



98 DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 63. Compound Indeterminate Forms. 

When a given function can be resolved into factors, one or more of 
which become indeterminate for a particular value of x, the true value 
may be obtained by getting the true value of each factor separately. 

When the true value of any indeterminate form is found, that of 
any constant power of it can be determined. 

PROBLEMS. 

lm l - s» ' Wh6n X = 1 ' 

X a ' 1 x c 

This may be put in the form 



l + x b l-x b ' 



in which the second factor only is indeterminate. 

Therefore [^CTJr^ 

2. (HLfH^B 9 when a = 0. 
X s 

(e*-l)tan 2 a; _ /tanaV (e x -l) 

X 3 \ X J X 



L x Jo L x 



therefore (e*-l)tan 2 x = - when x = 1 

XT 

3. log(l + « + <*) + 108(1-* + **) when „ = o. Ans. 1. 

sec a; — cos x 



(o On • 7TX 

x l — cr) sm 



4. 



2 a , when a; = a. Ans. 



ttX 

orcos?^ 



1 



x ( i^£) 3 ' when x= l' Ans ' ^ 



CHAPTER X. 

MAXIMA AND MINIMA OF FUNCTIONS. 

Art. 64. Definitions and Geometric Illustration. 

A maximum value of a function is a certain value at which, the 
function changes from an increasing to a decreasing function. Or, in 
other words, fix) is a maximum for that value of x which makes fix) 
greater than fix + h) and f(x — h) for very small values of h. 

A minimum value of a function is a certain value at which the 
function changes from a decreasing to an increasing function. Or, 
fix) is a minimum for that value of x which makes fix) less than 
f{x -f- K) and fix — h) for very small values of h. 

In Fig. 8, let the curve AB be the locus of y=f(x). 

Then PN represents a maximum ordinate, and P'T a minimum or- 
dinate. As x increases toward ON, y approaches a maximum value, 



y 




P 






/B 




A 






Ve 


P" 


p lv ^/ 




P™ 






fy^ 






A 










p 









f 


A I 


^1 f 


? < 


n 


-\ 


/ 






V 


V 


X 



Fig. 



PjV, and the tangent to the curve makes an acute angle with the 
X-axis. At the point P, the tangent line is parallel to the X-axis. 
Immediately after passing P, the tangent makes an obtuse angle with 
the X-axis. But by Art. 27, the slope of the tangent line is equal 



100 DIFFERENTIAL AND INTEGRAL CALCULUS. 

to f'(x) ; hence f'( x ) is positive before, and negative after, a maximum 
ordinate. Likewise it may be shown that f'(x) is negative before, and 
positive after, a minimum ordinate. Thus, f'(x) = 0, at both maxi- 
mum and minimum ordinates. Therefore, a condition for both maxima 

and minima is — = 0. 
dx 

Art. 65. Method of Determining Maxima and Minima. 

For a maximum value of the function, f'(x) = and f'(x) changes 
sign from + to — when x passes through a value corresponding to a 
maximum value of the function. For a minimum value of the func- 
tion, f'(x)=0 and f'(x) changes sign from — to + when x passes 
through a value corresponding to a minimum value of the function. 
Hence, the roots of the equation f'(x) = are first obtained. 

If a is a root of this equation, a value slightly less, and then one 
slightly greater than a, are substituted for x in f'(x). Let h repre- 
sent a very small quantity. 

If f'(a — h) is -f , and f'(a + h) is — , then f(a) is a maximum. 

If /'(a — h) is — , and f'(a + h) is +, then f(a) is a minimum. 

For example, let y = b + (x — a) 4 . 

Then — = 4 (x — a) 3 = ; hence x = a. 

dx K J 

Substituting a — h for x } gives 

^ = 4(tt-/*-a) 3 =-47i 3 . 
dx v J 

Substituting a -f h for x, gives 

^/ = 4(a + ft-a) 3 =+4A 3 . 
dx 

Here, /'(#) changes sign from — to + at x = a; hence, a is the 
value of x which gives a minimum function. Therefore, y = b, a mini- 
mum. 

By reference to Fig. 8, it will be seen that P'" IT is a minimum 
ordinate, and the tangent to the cuiwe at this point is perpendicular to 



MAXIMA AND MINIMA OF FUNCTIONS. 101 

the X-axis. In this case f'(x) changes sign by passing through oo . 
Any variable can change its sign only by passing through or oo , but 
it does not necessarily follow that there is a change of sign whenever 
f'(x) = 0, or f'(x)=ao. At point P IV , the tangent is parallel to the 
X-axis, hence f'(x) = 5 bvit f'( x ) is -f- immediately before and after 
reaching this value. Therefore, the values of x which make f'(x) = 
do not always give maxima and minima, so they are simply called 
critical values, or values for which the function is to be examined. 

It is evident also that a function may have several maxima and 
minima, and a minimum value may be greater than a maximum value 
of the same function. 

Art. 66. Conditions for Maxima and Minima by Taylor's 

Theorem. 

Let f(x) have a maximum or minimum value when x = a. 

Then if h be a very small increment of x, by Art. 64, 

f(a) >/(a + h), and f(a)>f(a — h), for a maximum, 
also f(a) <f(a + h), and f(a) </(« — h), for a minimum. 

Therefore /(a + h) —f(a) and f(a - h) —f(a) 
are each negative for a maximum, or are each positive for a minimum. 

Now by Taylor's Theorem, 

f(a + h) -/(a) = f'(a)h + /» | | 2 + /'"(a)|+ •••• (1) 

/(a-A)-/(a)=-/'(a)/i+/"(«)|-/"'(«)|+-- (2) 

For a maximum : The first members of (1) and (2) must be nega- 
tive, therefore the second members must be negative. Now if h be 
taken sufficiently small, the first term in each second member can be 
made numerically greater than the sum of all the terms following it; 
hence, the sign of each second member .will be the same as that of its 
first term. But the first terms have different apparent signs, so the 



102 DIFFERENTIAL AND INTEGRAL CALCULUS. 

second members cannot both be negative unless the first term disap- 
pears, hence 

/'(<*) = o. 

N~ow the first of the remaining terms of the second members con- 
tain h 2 , and these terms determine the signs of the members. In 
order that these terms may be negative, f"(a) mnst be negative, or 

Therefore, if f(a) is a maximum, 

f'(a) = and f"(a) < 0. 
Similarly, it may be shown that if f(a) is a minimum, 

f'(a)-0 and /"(a) > 0. 

However, if /"(a) = 0, then the sign of the second members of (1) 
and (2) will depend on the terms containing f'"(a), and since the terms 
containing /'"(a) have opposite signs, there can be neither a maximum 
nor a minimum unless /'"(a) also vanishes; and if f'"(a) — 0, then 
f(a) is a maximum when/ IV (a) is negative, and a minimum when/ IV (a) 
is positive ; and so on. 

Rule: Find f'(x), and solve the equation, f'(x) = 0. Substitute the 
roots of this equation for x inf"(x). Each value of x which makes f"(x) 
negative will make f(x) a maximum; and each value which makes f"(x) 
positive will make f(x) a minimum. 

However, if any value of x also makes f"(x)= 0, substitute this value 
in the successive derivatives until one does not reduce to 0. If this be of 
an odd order, the value of x will give neither a maximum nor a minimum; 
but if it be of an even order and negative, f(x) will be a maximum, if of an 
even order and positive, f(x) will be a minimum. 

The solution of problems in maxima and minima is often simplified 
by the aid of the following principles : 

1. If any value of x makes af(x) a maximum or minimum, a being 
a positive constant, that value will make f(x) a maximum or minimum. 
Hence, a constant factor may be omitted. 

2. If any value of x makes [/(#)] w a maximum or minimum, n being 
a positive constant, that value will make f(x) a maximum or minimum. 



MAXIMA AND MINIMA OF FUNCTIONS. 103 

Hence, any constant exponent of the function may be omitted; or if 
the function is a radical, the radical sign may be omitted. 

3. If any value of x makes logf(x) a maximum or minimum, that 
value will make f(x) a maximum or minimum. Hence, to find a maxi- 
mum or minimum value of the logarithm of a function, the function 
only need be taken. 

4. If any value of x makes fix) a maximum or minimum, that value 

will make a minimum or maximum. Hence, when a function is 

a maximum or a minimum, its reciprocal is a minimum or a maximum. 

5. If any value of x makes a + f(x) a maximum or minimum, that 
value will make fix) a maximum or minimum. Hence, a constant term 
may be omitted. 

Each of the preceding propositions may be readily proved. For 
example, in (1), the first derivative of af(x) when placed equal to zero, 
will give an equation whose roots are the same as the roots of the 
equation formed by placing the first derivative of fix) equal to zero ; 
hence, the critical values will be the same in both cases. 



PROBLEMS. 
1. Find the maximum and minimum values of 

Let y = x 3 — 3 x 2 — 9 x -f 5 ; 

then ^/ = 3^_6a-9. 

dx 

Placing the first derivative equal to zero, and finding the roots, 

3x 2 -6x-9 = 0-, 

therefore x = 3 or — 1. 

The second derivative is — = 6x — 6. 
dx 2 



104 DIFFERENTIAL AND INTEGRAL CALCULUS. 

When x = 3, — ^ = 12, and as this value of x makes the second de- 
dx 2 

rivative positive, it corresponds 

to a minimum value of the 

function. 

When 8=-l, ^/ = _ 12, 

dx 2 

and this result being negative 
indicates a maximum. 

Substituting these values of 
x in the function, gives, when 
x = 3, y = — 22, a minimum, 
and when x = — 1, y = 10, a 
maximum. . 

These results may be illus- 
trated graphically by construct- 
ing the locus of the equation. 

In Fig. 9 it will be seen that 
there is a maximum ordinate 
corresponding to the abscissa 
— 1, and a minimum ordinate 
corresponding to the abscissa 3. 
Remark. It will be very instructive to construct the loci of the 
equations in the first few examples. 

Examine the following functions for Maxima and Minima: 

Ans. x = 1, gives a Maximum, 2; 

x = 3, gives a Minimum, — 26. 




Fig. 9. 



2. y = X 5 — 5 8 4 + 5 X 3 + 1. 

3. ?/ = 2 a; 3 -21 a 2 + 36a-20. 

4. y = 3 x s - 9 x 2 - 27 8 + 30. 
(a - a;) 3 



5. 2/ 



a — 2a; 



^4?is. a; = l, gives Max., — 3; 

a; = 6, gives Min., — 128. 
Ans. x — — 1, gives Max., 45 ; 

8 = 3, gives Min., — 51. 

Ans. x — \a, gives Min., -fja 2 . 



MAXIMA AND MINIMA OF FUNCTIONS. 



105 



6. y = a; 3 - 3 x 2 + 6x + 10. 

7, y = 



x tan x 



8. y = xi. 

9. y = 



sina; 



1 + tan x 

10. ?/ = sin aj (1 + cos x). 

11. y=(*-.l) 4 (a? + 2)» 



.4/is. This function has no real Max. or Min. 
Ans. x = cos x, gives a Max. 

Ans. x = e, gives Max. 
,4ns. a; = 45°, gives Max. 

,471s. a; = y, gives Max. 

Ans. x = —-2-, gives Max. ; 
x = 1 3 gives Min. ; 
a = — 2, gives neither. 



GEOMETRIC PROBLEMS. 

12. Determine the maximum rectangle inscribed in a given circle. 
Assume an inscribed rectangle as in Fig. 10. Let the diameter 
CB = d, and the side CD = x ; then 



AC = -Vd 2 -x 2 . 
Denoting the area by A, then 



A — x Vd- — x 2 , 

which is to be a maximum. 

By Art. 66, 2, the function o?(d 2 — x 2 ) may 
be used. 




Fig. 10. 



Put 
Then 

Now 



y = xr(d 2 — X 2 ) 

dy 
dx 



2 d 2 x — 4 x 3 = ; hence x = 0, or x = d V+ 



dry 

dx 2 



= 2d 2 -12o: 2 = 2d 2 , when a; = 0; 
= — 4 d 2 . when x = d y/h 



Therefore, x = eJ-VJ? which is the side of an inscribed square, will 
give the maximum rectangle. 



106 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



13. Find the greatest cylinder which can be inscribed in a given 

right cone with a circular base. 

In Fig. 11, let CH be a cylinder 
inscribed in the cone OAB. 

Given AM = a, and OM = ft. 

Let NC = x, and NM= y. 

Denoting the volume by V, 

then V== irxhj. 

From the similar triangles AOM 
and COK 




Fig. 11. 



h h — v i a /T N 

- = ^-, hence x= -(h — y). 

ax h 



a 
Therefore, V= it — (h — y) 2 y, which is found to be a maximum when 

y — \li. Therefore, the altitude of the maximum inscribed cylinder is 
one-third of the altitude of the cone. 



14. Find the maximum cone which can be inscribed in a sphere 

whose radius is r. 

In Fig. 12, let ADB and CDB be 
the semicircle and triangle which gene- 
rate the sphere and inscribed cone by 
revolution about AB. 

Let CD = x, CB = y, and V= the 
□ volume of the cone ; 

then V=\irx-y. 

x 2 =CBx CA = y(2r-y), 
hence, V= ^rry 2 (2r — y), which is the function whose maximum is 




Fig. 12. 



required. 



Ans. The altitude of the Max. cone 



*r. 



15. Determine the right cylinder of greatest convex surface that 
can be inscribed in a given sphere. 

If r = the radius of the sphere, and x = the radius of the base of the 



MAXIMA AND MINIMA OF FUNCTIONS. 107 



cylinder, then the convex surface of the cylinder is 4 7rx-yV 2 — x 2 . This 

v 
will be a maximum when the radius of the base is 

V2 
16. From a given surface S, a cylindrical vessel with circular base 
and open top is to be made, so as to contain the greatest amount. 
To find its dimensions. 

Let x = radius of base, y = altitude, and V— volume of a cylinder. 

Then V= irx 2 y, (1) 

and S = ttx 2 + 2 nxy. (2) 

Differentiating (1) and (2) with respect to x : 

From (1), ^ = 2 wxy + irx 2 ( ^- = 0, 

dx dx 



hence 



dy = _2y, 
dx x 



dx 


_x + y m 

X 


2y 


x-+y 


X 


X 



From (2), = 2^ + 2^^ + 2 iry, 

dx 

hence 

Hence 

Therefore y = x, or the altitude = radius of base. 

In this example, (2) might have been solved for y and this value substituted in 
(1), and the solution would have been the usual one. But the given solution is 
in this and similar examples much shorter. 

17. What is the length of the axis of the maximum parabola which 
can be cut from a given right circular cone, knowing that the area of a 
parabola is equal to two-thirds of the product of its base and altitude ? 

Given BC = a, and AB = b, in Fig. 13. 

Let CM = x, then BM —a — x, 



and RS =2V(a-x)x. 



108 DIFFERENTIAL AND INTEGRAL CALCULUS. 



By similar triangles. 




Fig. 13. 
Hence, the area of the parabola is 

A = --x \/{a — x) x, 

which is a maximum when x = J a. 

18. What is the altitude of the maximum rectangle which can be 
inscribed in a given segment of a parabola ? 

O 




In Fig. 14, let BOO be the parabolic segment and AO = h. 
Let OH=x, 



MAXIMA AND MINIMA OF FUNCTIONS. 109 

then MH = V2px. 

Therefore, area of MBSJST = 2 -y/2pxQi - a;), 

which is a maximum when x = — 

3 

19. What is the maximum cylinder that can be inscribed in an 
oblate spheroid whose semi-axes are a and b ? 

Ans. Radius of base = -J- a V6 ; altitude = -§ b V3. 

20. Find the maximum right cone that can be inscribed in a given 
right cone, the vertex of the required cone being at the centre of the 
base of the given cone. Ans. The ratio of the altitudes is ^. 

21. What is the maximum isosceles triangle which can be inscribed 
in a circle ? Ans. An equilateral triangle. 

22. What is the altitude of the cone of maximum convex surface 
that can be inscribed in a sphere whose radius is 3 ? 

Ans. Altitude = 4. 

23. When is the difference between the sine and the cosine of any 
angle a maximum ? Ans. When the angle = 135°. 

24. If the strength of a beam with rectangular cross-section varies 
directly as the breadth, and as the square of the depth, what are the 
dimensions of the strongest beam that can be cut from a round log 
whose diameter is D ? Ans. Depth = D Vf. 

25. A rectangular box with a square base and open at the top, is to 
be constructed to contain 108 cubic inches. What must be its dimen- 
sions so as to contain the least material ? 

Ans. Altitude = 3 inches ; side of base = 6 inches. 

26. What is the altitude of the minimum cone that may be circum- v 
scribed about a sphere whose diameter is 10 ? Ans. Altitude = 20. 

27. A person, being in a boat 3 miles from the nearest point of the 
beach, wishes to reach in the shortest time a place 5 miles from that 
point along the shore ; supposing he can walk 5 miles an hour, but can 
row only at the rate of 4 miles per hour, required the place he must 

' land. Ans. One mile from the place to be reached. 



110 DIFFERENTIAL AND INTEGRAL CALCULUS. 

X s 4- 2 b 3 

28. Find the minimum value of y when y — — -• 

_ - 3 (a 8 + W») 

Arts, y = 0.32218. 

29. Determine the greatest rectangle which can be inscribed in a 
given triangle whose base = 2b and altitude = a. Ans. A = \ab. 

30. A Norman window consists of a rectangle surmounted by a 
semicircle. Given the perimeter, required the height and breadth of 
the window when the quantity of light admitted is a maximum. 

Ans. Radius of semicircle = height of rectangle. 

31. A privateer must pass between two lights A and B on opposite 
headlands, the distance between which is c. The intensity of light A 
at a unit's distance is a, and the intensity of light B at a unit's dis- 
tance is b. At what point must a privateer pass the line joining the 
lights so as to be as little in the light as possible, assuming the princi- 
ple of optics, that the intensity of a light at any point is equal to its 
intensity at a unit's distance divided by the square of the distance of 
the point from the light ? * _ cofi 

a? + b^ 

32. The flame of a candle is directly over the centre of a circle 
whose radius is 5 ; what ought to be its height above the plane of the 
circle so as to illuminate the circumference as much as possible, sup- 
posing the intensity of the light to vary directly as the sine of the 
angle under which it strikes the illuminated surface, and inversely as 
the square of its distance from the same surface ? 

Ans. Height above circle = 5 Vj. 

f 

33. If the total waste per mile in an electric conductor is OV + - 

r 

(r ohms resistance per mile), due to heat, interest, and depreciation, 
what is the relation between O, r and t when the waste is a minimum ? 

Ans. Cr = t. 

4 (x 2 4- x 4 ) rl 2 c 

34. In the formula, A 2 B = *—+ — — — '—, it is required to find the 

E 2 (x-iy 1 

x 2 4- x 4 
value of x that makes the variable factor - — — — - a minimum. 



Ans. a = 2.2. 



MAXIMA AND MINIMA OF FUNCTIONS. Ill 

35. From the differential equation, 20,000,000 ^11 = - 100 x, find 

dor 

the equation of the curve and the maximum ordinate ; the first con- 
stant of integration being found by making -J- = when x equals I, 

dx 

and the second constant of integration being found by making y equal 
to zero when x equals zero. 

A 1 /lOOfo 100 oA ,.„ r , I s 

Ans. y = , and Max. ordinate = 

J 20,000,000 V 2 6 / 600,000 

36. A statue whose height is 10 feet stands on a pedestal 8 feet in 
height, which rests on a level surface. At what point on the horizontal 
plane through the base of the pedestal does the statue subtend the 
greatest angle ? Ans. 12 feet from centre of base. 

37. If v denotes the velocity of a current, and x the velocity of a 
steamer through the water against the current, what will be the speed 
most economical in fuel if the quantity of fuel burnt is proportional 
to X s ? Ans. x = &v. 



Art. 67. Maxima axd Minima of Functions of Two Indepen- 
dent Variables. 

Let f(x, y) represent any function of two independent variables. 

When f{x, y) >f(x + h, y + 7c), 

for all very small values of h and k, positive or negative, the function 
has a maximum value. 

When f{x, y) <f(x + h, y + 7c), 

for such values of h and 7c, the function has a minimum value. 

Placing u =f(x, y), 

from Art. 57, 

f(x + h,y + k) -fix, y) = h^L + k^ 

ox oy 



+,' 



h — - + 2 hk — — - + k- —— 
dx~ oxoy oy 



(i) 



Now, by argument similar to that of Art. 66, it may be proved that 
'the sign of f(x + h, y -\-7c) —fix, y) will depend on h t~ + &-t-j an( ^ 



112 DIFFERENTIAL AND INTEGRAL CALCULUS. 

therefore will change sign with h and k ; hence, a maximum or mini- 
mum value is impossible unless 

dx dy 
And since h and k are independent, 

— = 0, and — = 0. (2) 

dx ' dy K J 

o ?y etui fj u 

Substituting -4 = — -, jB = — — , and (7= — -, in equation (1), gives 
dx 1 dxdy by 1 

fix + h,y + k)- f(x, y) = i (Ah 2 + 2 5M + Gk 2 ) + • . . 



IF 
12 4 



4| + sY+(4(7-£ 2 )~U... (3) 



In (3), the sign of f(x + h, y + ft) —f(x, y) will depend on 



^l+^Yi ! 



and in order that it may retain the same sign for all very small values 
of h and k, it is necessary that AG — B 2 should be positive; for if 
AC — B 2 be negative, it will be possible, by ascribing some suitable 
value to - to make the whole expression change its sign. Hence as a 

a, 

condition for a maximum or minimum, 

B 2 <AG (4) 

It is obvious from (4) that A and G will have the same sign, and 
the sign of (3) thus depends on A. 

Hence, for a maximum, A < 0, and G < ; 

and for a minimum, A > 0, and C > 0. 

Therefore, the conditions established are : 
For either a maximum or minimum, 

** = 0, *f = 0, and (J^LY^Shi 

dx dy \dy dx) dx~ dy 2 



MAXIMA AND MINIMA OF FUNCTIONS. 113 

Also, for a maximum, — - < and — - < 0, 

dx 2 dy 2 

and for a minimum, — - > and — - > 0. 

dx 2 dy 



art. 68. conditions for maxima and mlnima of functions of 
Three Independent Variables. 

By an investigation similar to that of Art. 67, the following condi- 
tions for a maximum or minimum value of u =f(x, y, z) are established : 
For either a maximum or minimum, 

du_r, 5w_a 5^_o ( dhi V d 2 u d 2 u 



dx dy dz ' \dx By) dx 2 dy 2 ' 

J^Y<^, and (*>.)•<***». 

dy dzj dy dz 2 \dz dx) dz- dx 2 



d'U ^ r, dhi ^ A -, d 2 u 

— ^ < 0, — < 0, and — 
dx- dy 2 dz- 



Also, for a maximum, — < 0, -^ < 0, and -^ < 0, 



and for a minimum, — - > 0, — - > 0, and — - >0. 

dx- dy- dz 2 



PROBLEMS. 

1. Find a point so situated that the sum of the squares of its dis- 
tances from the three vertices of a given triangle shall be a minimum.* 

Let (<&,, y x ), (x 2 , y 2 ) and (x 3 , ?/ 3 ) be the coordinates of the vertices, 
and (x, y) the given point. 
Then, 

l(x - x,) 2 + (y - Vl J] + [0 - x 2 ) 2 + (y- y-n + l(x - x s ) 2 + (y - y s ) 2 ] 

is the function to be a minimum, which may be represented by u. 
^=2(x-^x 1 ) +2(x-x 2 ) +2(x-x 3 ), 

^ = 2(y-y 1 ) + 2(y-y 2 )+2( : y-y 3 ), 
* See Byerly's Biff. Calc, p. 236. 



114 DIFFERENTIAL AND INTEGRAL CALCULUS. 

^=2 + 2 + 2 = 6 = ^, 
dx 2 

dHl = B, 



dxdy 

^=2 + 2 + 2 = 6 = 0. 
dy 2 

Making — and — each equal to zero : 
dx dy 

2(x- xd + 2 (x — x 2 ) -f- 2 (x - x 3 ) = 0, 

therefore x = Xl + x * + x * - 

3 

2 (2/ - yi ) + 2 (2/ - 2/ 2 ) + 2 (2/ - 2/ 3 ) = 0, 

therefore y = Vx + y2 + ^ 3 - 

* 3 

^LO- £ 2 = 36 -0>0, 

A = 6 > 0. 
Hence, w is a minimum when 

a , = «i + ^ + aii )andy = y. + yi + y, ) 

and the required point is the centre of gravity of the triangle. 

2. Find the maximum value of x s y 2 (6 — x — y). 

Ans. Max. when x = 3, y = 2. 

3. Find the maximum value of 3 axy — x 3 — y 3 . 

Ans. Max. when x = a, y = a. 

4. What is the triangle of maximum perimeter that may be in- 
scribed in a given circle ? . . . 

Ans. An equilateral triangle. 

5. Find the values of x, y and z, that make x~ -\- y 2 -f 2 2 + x — 2 2 

— aw a minimum. 

^4)is. x = — f , ?/ — — h z = !• 



MAXIMA AND MINIMA OF FUNCTIONS. 115 

6. What rectangular parallelopiped inscribed in a given sphere has 
the maximum volume ? Ans. A cube. 

7. An open vessel is to be constructed in the form of a rectangular 
parallelopiped, capable of containing 108 cubic inches of water. What 
must be its dimensions to require the least material in construction ? 

Ans. Length and width, 6 inches ; height, 3 inches. 



CHAPTER XI. 

TANGENTS, NORMALS AND ASYMPTOTES TO ANY PLANE CURVE. 

Art. 69. Equations of the Tangent and Normal. 

Let y =f(x) be the equation of any plane curve AB, and (V, y') the 
coordinates of the point of tangency T, in Eig. 15. The equation of a 




Fig. 15. 



straight line through T is y — y' = a(x — x'), in which a is the tangent 
of the angle which the line makes with the X-axis. 



By Art. 27, 



Therefore 



a = tan \b = — • 
dx' 

y — if = -Z-(x — x') 



(1) 



is the equation of the tangent to the curve y =/(#) at the point (V, y'). 

As the normal to the curve at the point T is perpendicular to the 

dx' 
tangent at that point, the slope of the normal is — — -, and hence the 

dy' 
equation of the normal is 



y -y 



dy' K J 

116 



(2) 



TANGENTS, NORMALS AND ASYMPTOTES. 117 



Art. 70. Lengths of Tangent, Normal, Subtangent and 

Subnormal. 

In Pig. 15, let TM be the- ordinate, TE the tangent and TN the 
normal, at the point of contact ; then ME is the subtangent and MN 

the subnormal. 

„ ir TM ,dx' 

RM — = y' — : 

tan MET dy'' 

dx' 

hence subtangent = y' (3) 

6 9 dy' v J 

MN= MTtamMTJSr= i/^-\ 

dx' 

dv' 
hence subnormal = y'—; (4) 

9 dx' y } 



ET = VWW 2 + (MTy = yjfy'^y+ y* 



hence tangent = y\ll -f-( — ] • (5) 

V \dy'J 

TN=^(MTy+(MN)* = yjy's + fy'^i • 
hence normal = y\l -f ( ^ ) ■ (6) 



Art. 71. Tangent of the Angle between the Radius Vector 
at Any Point of a Plane Curve and the Tangent to the 
Curve at that Point, in Polar Coordinates. 

Let be the pole, OX the initial line, and P any point of the curve 
AB, in Pig. 16. 

Let (r, 6) be the coordinates of P, and (r + Ar, + A0) be the 

coordinates of another point E of the curve. If PS is perpendicular 

to OE, then 

PS = r sin &6, 

and SE = (r + Ar) — r cos A0. 

Therefore tan SEP = rsinAfl 

r + A?- — r cos A0 



118 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Now, if the point R approaches P, then the secant RP approaches 
the tangent PT, and the angle SRP approaches the angle OPT. If 
the angle OPT is represented by </>, then 




Fig. 16. 



tan <f> = limit 



= limit 



sinA0 



r -\- Ar — r cos A0 
r sin A0 



A0 



r snr 





2 Ar 
Ad A6 


XT v .,sinA0 ., t ., Ar clr 
Now, limit = 1 : limit — = — - ; 

' AO Ad ae y 


o • 2 A6> 
2sm 2 — 

limit —_ = limit 

AO 


. A0 

sm — 

2 . AO A 
sm — ■ — 0. 
A0 2 




2 


Therefore tan <b = r — 

dr 





(1) 



Art. 72. Derivative op an Arc. 

In Fig. 17, let P and P' be two points on the curve AB separated 
by a short- distance As. The coordinates of P and P' are (x, y) and 

(x + Ax, y + Ay) respectively. 



TANGENTS, NORMALS AND ASYMPTOTES. 



119 



In the right triangle PMP\ 



hence 



A6'=V(A^) 2 + (A?/) 2 ; 



As 

Ax 



-v^i 





Y 


s/ 

A 

X 


F 

y 


AX 


fiy 

M 


B 















X 



Fig. 17. 



Now, when P' approaches P, 



limit — = limit 
Ax 



May- 



dx \ 






Art. 73. Derivative of an Arc in Polar Coordinates. 

Prom Fig. 16, regarding the limiting triangle of PSR, 

PR As 

limit sec PRS = limit —— = limit — > 

RS Ar 



hence 



therefore 



sec d> = — 
dr 

ds 



tl d0 



™ = Vl + tan 2 <£ =a/1 + ?' 2 ( t 1 ,by Art. 71, (1) 
c?s c?s dr 



L (dr 
dO drdO V [dO 



(1) 



(2) 



120 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 74. Lengths of Tangent, Normal, Subtangent, Subnormal 
and Perpendicular on Tangent, in Polar Coordinates. 

In Fig. 18, let PT be the tangent to the curve at the point P, and 
through P the normal PN is drawn. 




NT is drawn through the pole 0, perpendicular to OP. 
PT is called the polar tangent; 
PN, the polar normal; 
OT, the polar subtangent; 
ON, the polar subnormal; 
and OD, the perpendicular on tangent from the pole. 

OT= subtangent = OP tan OPT = Op(r c ^f], by Art. 71, 



V dr 



= r 2 



dO 
dr 



ON= subnormal = OP tan OPN= OP cot OPT 

_ dr 
~d$ 



(1) 



(2) 



TANGENTS, NORMALS AND ASYMPTOTES. 121 

PT = tangent = VOP 2 + OT~ = ryll + i 



dr 



\cie 

OD = perpendicular = OP sin OPD 

tan OPD 



VI + tan 2 OPD 2 , fdr\* 



PROBLEMS. 



1. Find the equations of the tangent and normal to the circle, 



x 2 + y 2 = r 2 . 



(3) 



PN = normal = V OP 2 + ON 2 = ^r 2 + (~ Y- (4) 



(5) 



By differentiation ^ = - - ; .-. ^ = - *'. 

Substituting this derivative in Art. 69 (1), gives 

y-y' = - -{? ~ x '^> 
whence, xx' -f yy' = r 2 , which is the equation of the tangent. 

Substituting — , = in Art. 69 (2), gives y — y'= —(x — x'), 

dx y' x 

which is the equation of the normal. 

2. Find the equations of the tangent and normal to the ellipse 

tftf + b 2 x 2 = a 2 b 2 . 

3. Find the equations of the tangent and normal to the parabola 

y 2 = 2px. 

4. Find the equations of the tangent and normal to y 2 = 2x 2 — x 
at x == 1. Ans. Equations of tangent : y = ±x + ±, y = — ±x — \. 

Equations of normal : y = — 2 x + 3, y = 2x — 3. 

5. What is the inclination of the tangent to the curve 

x^y 2 = a s (x + y), at the origin ? Ans. 135°. 

6. What is the value of the subtangent to y = a x ? Ans. m. 



122 DIFFERENTIAL AND INTEGRAL CALCULUS. 

7. What is the value of the subnormal to y n — a n ~ x x ? Ans. 



f 



8. At what angle do the curves y = — — - — - and x 2 — kay intersect ? 



nx 

8 a 3 

x 2 -|- 4 a 2 



-4ws. 71° 33' 54". 

9. Find the values of the normal and subnormal to the cycloid 

x — 2 arc vers - — V4 y — ?/ 2 , at the point where y = 1. 

^4ws. Normal = 2 ; subnormal = V3. 

10. Find the values of the tangent, normal, subtangent and sub- 
normal to the spiral of Archimedes, whose equation is r = ad. 

Ans. Subtangent == - , subnormal = a, 
a 



tangent = r\l + — , normal = -y/r 2 + a 2 . 
* a 2 

11 . Find the values of the subtangent and subnormal to r 2 = a 2 cos 2 0. 

7* 



Ans. Subtangent = 



a 2 sin 2(9' 



subnormal = sin 2 6. 

r 



Art. 75. Rectilinear Asymptotes. 

An asymptote to a curve is the limiting position of the tangent 
when the point of contact moves to an infinite distance from the 
origin. 

Hence, any curve will have an asymptote when the point- of con- 
tact of a tangent is infinitely removed from the origin, and when the 
tangent intersects either coordinate axis at a finite distance from the 
origin. 

From Art. 69 (1), 

Intercept on X=x'-y' —. (1) 

dy' 

and intercept on Y=y' — x'-^-- (2) 

1 y dx' } 



TANGENTS, NORMALS AND ASYMPTOTES. 123 

If in (1) and (2) the intercept on X or Y is finite when x' — co or 
y' = oo, then the tangent at (x' } y') is an asymptote. For example, to 
examine the hyperbola cry- — b 2 x* = — crb 2 for asymptotes : 

XT cly' b 2 x' 

Here *,= —,' 

ax cry 

a 2 v' 2 
Hence intercept on X = x' — — ~^— 

1 b 2 x' 

= — = 0, when x' = oo. 
a 1 ' 

Intercept on Y= y' f- = = 0, when y' = oo. 

Hence, there is an asymptote passing through the origin. 

cly' b 2 x' b 1 6 , 

-7— ,= -o— ,= ± , = ± -, when x = 00. 

Therefore, there are two asymptotes whose slopes are ± -, and 
the equations of the asymptotes are y = ± - x. 

If, when x' = co in (1) and (2), the intercepts on both X and Fare 
infinite, the curve has no asymptote corresponding to x' = 00. 

If when y' = oo in (1) and (2), the intercepts on X and Fare 
infinite, the curve has no asymptote corresponding to y' = co . 

If both intercepts are zero, the asymptote passes through the origin, 

and its direction is found by evaluating -2 for x = 00. 

dx 

Art. 76. Asymptotes Parallel to an Axis. 

When x = 00 in the equation of a curve gives a finite value of y, 
then there is an asymptote parallel to the X-axis. For instance, if 
y = a when x = co in the equation of the curve, then y = a is the equa- 
tion of an asymptote, because it is the equation of a straight line 
passing within a finite distance of the origin, and touching the curve 
at an infinite distance. 

Likewise, when y = 00 gives x = b in the equation of a curve, then 
x = b is an asymptote parallel to the I r -axis. 



124 DIFFERENTIAL AND INTEGRAL CALCULUS. 

x 3 



For example, taking the curve whose equation is y 



x — b 
Here, y = oo when x = b ; hence x — b = is the equation of an 

asymptote to the curve parallel to the F-axis. 

Art. 77. Asymptotes determined by Expansion. 

An asymptote may sometimes be determined by solving the equa- 
tion of the curve for y and expanding the second member into a series 
in descending powers of x. 

x 3 
For example, to examine if 



x — b 

Here y 2 = x 9 ' 

x 



hence y = ± x f 1 — 



b\-l 



<c 



= ±x 
as x approaches x, (1) approaches 



^2a: Sar^lCr'^ J w 



2/=±(W|). (2) 

Hence, as x increases, the curve (1) is continually approaching the 
straight line (2), and (1) and (2) become tangent when x — oo ; there- 
fore, y = ± ( x -f- ] are the equations of two asymptotes to the curve (1). 



Art. 78. Asymptotes in Polar Coordinates. 

If a polar curve has an asymptote, as the point of contact is at an 
infinite distance from the pole, and as the tangent line passes within a 
finite distance from the pole, the radius vector of the point of contact 
is parallel to the asymptote, and the subtangent is perpendicular to the 

asymptote and finite. [See Fig. 18, Art. 74.1 

dO 

Hence, for an asymptote, the polar subtangent r 2 — is finite for 

r = op. Therefore, to examine a polar curve for an asymptote, a value 
of is found which makes r = oo ; if the corresponding polar subtan- 



TANGENTS, NORMALS AND ASYMPTOTES. 



125 



gent is finite, there will be an asymptote, and if the subtangent is in- 
finite, there is no corresponding asymptote. 

For example, to examine the hyperbolic spiral rO = a for asymp- 
totes. . 

If r = oo in r = -, then = 0. 
6 

dO a , odO 

— = -, hence r — = — a. 

dr v dr 

Therefore there is an asymptote parallel to the initial line which 
passes at a distance a from the pole. 



PROBLEMS. 

Examine the following curves for asymptotes : 

Arts. Asymptote, y = — x. 

Ans. No asymptotes. 



1. y 3 = a 3 — x 3 . 

2. The circle, ellipse, and parabola. 

3. y 3 = ax 2 + x 3 . 
x 3 



4. The cissoid, y 2 

5. y 3 = 2 ax 2 — x 3 . 

o 3 

6. y = c-h 



'Zr — x 



a° 



(x-bf 
7. The lituus, r0* = a. 



Ans. y = x -j — 

Ans. x = 2r. 

Ans. y = — x + -| a. 

Ans. y = c, and x = b. 

Ans. The initial line. 



8. ?'cos = a cos 2 0. 

Ans. There is an asymptote perpendicular to the initial line at 
a distance a to the left of the pole. 



CHAPTER XII. 



DIRECTION OF CURVATURE. POINTS OF INFLECTION. 
OF CURVATURE. CONTACT. 



RADIUS 



Art. 79. Direction of Curvature. 

A curve is concave towards the X-axis at any point, when in the 
immediate vicinity of that point it lies between the tangent and the 
X-axis. A curve is convex towards the X-axis when the tangent lies 
between the curve and the axis. 



Y 







p 








A 


V' 


y" 

h 




B 













X 



Fig. 19. 



In Fig. 19, let the coordinates of P be (x\ y'). The curve being 
concave downward, the ordinates of the curve for the abscissas x' ± h, 
h being a very small quantity, must be less than the corresponding 
ordinates of the line tangent to the curve at P. 

Likewise, in Fig. 20, the curve being convex downward, the ordi- 
nates of the curve for the abscissas x' ± h must be greater than the 
corresponding ordinates of the tangent line at P. 

126 



DIRECTION OF CURVATURE. 



127 



In either case, let (x' + h, y") be the coordinates of P'. 

If y =f(x) is the equation of either curve, then y" = f(x' -f- h), and 

y" =f(x l + K) =/(*') +f'(x')h +f"(x')^+f" 0O.| + - •■ 




Fig. 20. 



The equation of the tangent at P is 

y-y r =f'(x t )-(x-x'). 

If the coordinates of P 1 are (x' + h, y 2 ), 

y 2 -y'=f'(x')[(x' + h)-x'l 
Now y 2 =/(*') +/'<V)fc; 

hence 



/'-&=/"(*'>!+/'" (*')j!+ 



or if 7i be taken sufficiently small, 



(1) 

(2) 
(3) 



In (4), y" — y 2 will have the same sign as/" (V) ; therefore, the 
curve is concave to the X-axis if f"(x') is negative, and convex if 
f"(x') is positive. 

If the curve is below the X-axis, y" and y 2 are negative, and the 
curve is convex towards the X-axis when — y" + y 2 is negative, that is, 



128 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



if /"(#') is negative, and the curve is concave towards the X-axis when 
/" (x') is positive. 

Art. 80. Direction of Curvature in Polar Coordinates. 



A curve referred to polar coordinates is concave to the pole at any 
point, when in the immediate vicinity of that point it lies between the 
tangent to the curve at that point and the pole. A curve is convex 
to the pole when the tangent lies between the curve and the pole. 

If p is the perpendicular distance from the pole to the tangent to 
the curve at a point whose coordinates are (r, 0), it is evident from 
Fig. 21, that when the curve is concave to the pole, p increases as r 

increases : hence, -±- is positive. 
dr 




Fig. 21. 



Fig. 22. 



Similarly, from Fig. 22, when the curve is convex to the pole, p 

decreases as r increases ; hence -±- is negative. 

dr 

If the equation of the curve is given in terms of r and 6, the equa- 
tion may be transformed into an equation between r and p by aid of 

Art. 74 (5) • then the curve is concave or convex, according as — is 

dr 

positive or negative. 



POINTS OF INFLECTION. 129 



Art. 81. Points of Inflection. 

A point of inflection of a curve is a point where the curvature 
changes from concavity to convexity or the reverse. Hence, at a point 
of inflection the curve cuts the tangent. 

By Art. 79, when /" (x) < 0, the curve is concave to the X-axis, and 
convex when /"(#)>0; therefore f"(x) changes sign, and hence 
f"(x) = or oo at a point of inflection. 



x* 






For example, to examine y = — for points of inflection. 

d 2 y 6x — 2a fi - l a 

-4 = ; = : hence x = — 

dx 2 b 2 ' 3 

If x<% thenf^<0; 

3 dx 2 

and if x>-, then ^ o >0. 

3' dx 2 

Hence, f"(x) changes sign at the point whose abscissa is -, and 
therefore this will be a point of inflection. 



PROBLEMS. 

1. Find the direction of curvature and point of inflection of 
y = a + (x — b) 3 . 

Ans. There is a point of inflection at (a, b) ; on the left of this 
point the curve is concave, while on the right it is convex. 

2. Examine y = x + 36 x 2 — 2 X s — x A for points of inflection. 

Ans. At x = 2, and x = — 3. 

X s 

3. Find points of inflection and direction of curvature of y = 

1 * x 2 +12 

Ans. (— 6, — |), (0, 0), (6, f); convex on the left of the first point, 
concave between first and second points, convex between second and 
third, and concave on the right of third point. 

4. Find the direction of curvature of the lituus r = — 

Ans. Concave towards the pole when r < a^/2 • 
convex towards the pole when r > a^/2. 



130 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 82. Curvature. 

The total curvature of a curve between two points is the total 
change of direction in passing from one point to the other, and is 
measured by the angle formed by the tangents to the curve at the two 
points. 

The actual curvature of a curve at a given point is the rate of 
change of its direction relative to that of its length.* 




Fig. 23. 

In Fig. 23, let P be any point of the curve AB. The angle 
PTX = i//, or the angle which the tangent at P makes with the X-axis 
is the direction of the curve at P. Likewise the angle P'SX is the 



* Leibnitz defined the curvature of a curve at any point as the rate at which 
the curve is bending, or the rate at which the tangent is revolving per unit length 
of curve. 



As ~ 


1. 

= j 
r 


dxjj _ 
ds 


_1 
r 



RADIUS OF CURVATURE. 131 

direction of the curve at P. Then angle TCS = Ai// is the difference 
of these inclinations, and if PP' = As, and the point P' approaches P, 

limit ^! = # 

As ds 

which is an expression for the curvature. 

If the curvature is uniform ; that is, if AB is the arc of a circle 
whose radius is r, the angle TCS = angle PMP' at the centre subtended 

by the arc PP', 

PP' 

and angle TCS — arc ; 

hence 

therefore 

ds r 

Hence, the curvature of a circle is equal to the reciprocal of its 

radius. 

Art. 83. Radius of Curvature. 

The curvature of a circle varying inversely as its radius, and as 
any value at pleasure may be given to the radius, it follows that there 
is always a circle whose curvature is equal to the curvature of any 
curve at any point. The circle tangent to a curve at any point and 
having the same curvature as the curve at that point is called a circle 
of curvature ; its centre is the centre of curvature, and its radius is the 
radius of curvature at that point. 

Denoting the radius of curvature by p, by Art. 82 (1), 

ds n 

Now it is required to find p in terms of x and y. 

By Art. 27, tan^ = ^; 

dx 

hence sec 2 d/d\l/ = — - ; 

dx 

d 2 y cPy 

therefore # = _^L = ^_ (2) 

sec> ±4_{dM\ 

\dx 



132 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Substituting in (1) the value of d\p just found and 



4 



dy\ 



ds = dx\l + pM from Art. 72, 



ds 

d4> 



dxj 

Mil 

dx 2 



(3) 



which is the required radius of curvature. 

If I represents the length of the curve, equation (3) may be reduced 

to the formula p = — . 

dx • dry 



Art. 84. Radius of Curvature in Terms of Polar Coordinates. 

Formula (3), Art. 83, is first transformed, any quantity t being taken 
as the new independent variable. The values of -^ and — ^ from (2) 

(XX ClQb 

and (3) of Ex. 3, Art. 50, being substituted in (3) of Art. 83, putting 
t = 0, 



P = 



dx 2 dy 2 

do 2 ae 2 

dx 2 
dO 2 

fdx 2 d 
[dO 2 d 


2 d 2 y dx d 2 x dy 
d0 2 "d0~d0 2 d0 


dx s 
d& 

e 2 ) 



d 2 y dx _ d 2 x dy 
d& 2 'dV~d(P'dO 



(1) 



From the equations of transformation, x — r cos 0, and y = r sin 0, 
by differentiation, 



dx _ 
d0~ 


dv 

- r sin + cos 6 — , 
dO 






dy_, 

dO 


dv 

- cos + sin — , 
dO 






d 2 x_ 
dO 2 


- r cos 6 — 2 sin — + 
dd 


cos 




d 2 y _ 
db 2 ~ 


dr 
- r sin + 2 cos 6 ^ + 
eft? 


sin 


a d 2 r 



CONTACT. 133 

Substituting these values in (1), 



P = 



dO 2 



' (16 dff- 



(2) 



which is the required formula. 



Art. 85. Contact of Different Orders. 

Let y =f(x) and y = <j> (x) be the equations of any two curves re- 
ferred to the same axes. 

Giving to x a small increment h, and expanding, 

/(■ + ft) = f(x) +/'(»)* + f"(x) |+ /'"(*) |* + »., (1) 

+ (» + ft) = + (s) + +'(»)* + < ^»( a! )|+^»(x)j|+.... (2) 

If the two curves have a common point whose abscissa is a, then 
f(a) = <f> (a). If, furthermore, /'(a) = <£'(a), the curves have a com- 
mon tangent ; this is called contact of the first order. 

If, also, f"(x) = <£"(#)> the two curves have contact of the second 
order. 

In general, two curves will have contact of the ?ith order at x = a, 
when the following conditions are satisfied : 

f(a) = $ (a), /'(a) = *'(a), /"(a) = <£"(a), - f n (a) = 0»(a). 

If the curves have a common point at # = a, and if a be substituted 
for x in (1) and (2), and (2) be subtracted from (1), then 

/(« + A) _ + (a . + h) = ft[/'(o) - $'(a)]- £ [/"(a) - +»(a)] 

If 

+ |[/"'(«) .- *"'(«)] + -, (3) 

which is the difference between corresponding ordinates of the curves. 

Now, if these curves have contact of the first order, the first term of 

the second member of (3) reduces to zero ; if they have contact of the 

second order, the first two terms reduce to zero; and so on. Hence, 



134 DIFFERENTIAL AND INTEGRAL CALCULUS. 

when the order of contact is odd, the first term which does not reduce 
to zero contains an even power of h, and the sign of the second member 
is the same whether h be positive or negative ; therefore, one curve lies 
above the other on each side of their common point, and the curves do 
not intersect. But when the order of contact is even, the first term 
which does not vanish contains an odd power of h, and in this case the 
second member changes sign with h; therefore, one curve lies above 
the other on one side of the common point, and below it on the other 
side, and the curves intersect. 

Art. 86. Radius of the Osculating Circle, and Coordinates 

of its Centre. 

It appears from Art. 85, that n + 1 equations must be satisfied 
when a curve has contact of the nth. order with another curve. As 
an equation may be made to conform to as many different conditions 
as there are arbitrary constants in it, it follows that the number denot- 
ing the order of contact which any curve may have is one less than the 
number of arbitrary constants in its equation. The general equation 
of the circle has three constants ; hence, at any point of a curve, the 
circle will have, in general, contact of the second order ; this circle is 
called the osculating circle. 

Let the equation of the given curve be 

y =/(*), (i) 

and the equation of the circle 

(x'- a y + (y'-b) 2 = r 2 . (2) 

Differentiating (2) twice, 

*'-a+<y-&)|J = 0, (3) 

and i+^+y-^g-a (4) 

If (2) is the osculating circle at the point (x, y) of (1), 

x' = x and y' = y, ^ = % ^1 = ^1. 
dx' dx dx L dx 2 



CONTACT. 135 

Substituting these values in (4), and solving for b : 



From (3), a = x — 



dx 2 



2~l 



da? 2 



dy 
dx 



(6) 



Substituting values of (y — b) and (as — a) from (5) and (6) in (2), 
after reducing, 

r = ± l WJ, (7) 

The values of a and 6 in (5) and (6) are the coordinates of the 
centre of the osculating circle, and the value of r in (7) is its radius. 
Hence, by comparison with Art. 83, it will be seen that the osculating 
circle is the circle of curvature and the radius of the osculating circle 
is the radius of curvature. 

Art. 87. The Osculating Circle has Contact of the Third 
Order where the Radius of Curvature is a Maximum or 
Minimum. 

If p is to be a maximum or minimum, by Art. 66, 

dx 
Differentiating (3) of Art. 83, 

do 2{ ^da*J * dx\drfj dx\ ^dx*J 

\dx 2 ) 
s ty/d?y s 

therefore ^= C? ^f / ■ 

^dx 2 



136 DIFFERENTIAL AND INTEGRAL CALCULUS. 

The same value of — | as found in (2) will be obtained by differen- 
ctx 

tiating (5) of Art. 86. Therefore, the given curve and the osculating 

(Pi i 

circle have the same value for — - at a point of maximum or minimum 

dx 3 

curvature ; hence the contact at such a point is of the third order. 



PROBLEMS. 



1. Find the radius of curvature of the parabola y 2 = 2px at any 
point, and the coordinates of the centre of curvature. 



dy __p d 2 y _ p 2 . 



dx y dx 2 y 3 

hence , = ± ^L = (£+M = £ 

_ Pi P P 



_„ L t 



a = x 



p 2 



= Sx +p. 



1 



p 2 



f 

At the vertex, x = and y = ; therefore p= p, a =p and 6 = 0. 

2. Find the radius of curvature of the ellipse a 2 y 2 + b 2 x 2 = a 2 b 2 . 

3. Find the curvature of y = x 4 — 4 X s — 18 x 2 at the origin. 

^ws. p = 36. 

4. Find the radius of curvature of the cycloid x = r arc vers - 



V2 r?/ — y'i Ans. p = 2 V2 r?/. 



PROBLEMS. 137 

X X 

5. Find the radius of curvature of the catenary y = - (e a + e a ). 

Ans. p = — • 
r a 

6. Find the radius of curvature of the spiral of Archimedes, r = aO. 

Ans. !£±*± 

2 a 2 + r 2 

7. Find the radius of curvature of the cardiod r = a (1 — cos 0). 

Ans. p = f V2 ar. 

8. Find the radius of curvature of the ellipse whose axes are 8 and 
4, at x = 2, and the coordinates of the centre of curvature. 

Ans. P = 5.86 ; a = .38, and b = - 3.9. 

9. Find the order of contact of 

x 3 — 3 x 2 = 9 y — 27 and 3 # = 28 — 9 y. Ans. Second order. 

10. What is the order of contact of the parabola 4?/ = x 2 — 4 and 
the circle x 2 -\-y 2 — 2y = 3? Ans. Third order. 

11. What is the radius of curvature of the curve 16 y 2 = 4 x 4 — x s , 
at the points (0, 0) and (2, 0) ? Ans. p = l, and p = 2. 

12. What is the radius of curvature of the curve y = x 3 + 5 x 2 + 6 a?, 
at the origin ? ^ws. p = 22.506. 

13. Find the radius of curvature and the coordinates of the centre 
of curvature of the curve y = e x , at x = 0. 

-4ns. p = 2V2, (a, &) = (- 2, 3). 



CHAPTER XIII. 



EVOLUTES AND INVOLUTES. ENVELOPES. 

Art. 88. Definition of Evolute and Involute. 

The evolute of any curve is the curve which is the locus of the 
centres of all the osculating circles of the given curve ; the given curve 
with respect to its evolute being called an involute. 




Fig. 24. 

Iii Tig. 24, let AB be the given curve, and the centres of curvature 
of P', P", P'", etc., be respectively P lf P 2 , P 3 , etc. ; then the curve MN, 
which is the locus of P ly P 2 , P 3 , etc., is the evolute of AB. 

Art. 89. Equation of the Evolute. 

The equation of the evolute is the equation which expresses the 
relation between the coordinates of the centres of all the osculating 

138 



/ 



EVOLUTES AND INVOLUTES. 



139 



circles of the involute. The values of -^ and — \ derived from the 

dx dx 2 

equation of the curve are substituted in equations (5) and (6) of Art. 
86, giving two equations, which, together with the equation of the given 
curve, make three equations involving x, y, a and b ; by combining these 
equations, eliminating x and y, a resulting equation will be obtained 
showing a relation between a and b, the coordinates of the evolute, 
which is the required equation. 




Fig. 25. 



For example, to find the equation of the evolute to the common 
parabola, y 2 = 2px. 
, Here <*L = l,il = -£. 



dx y dx 2 y 2 "' 

Substituting in (5) and (6) of Art. 86 : 

&= i+? ; tf hence f=p i b i 

* y 2 p 2 p l £ 

y 2 _|_ p l p y& 

a = x-{- - — f— • - • Z-s = 3 x -+- r> ; hence x = 
y 2 y p 2 



a —p 



(2) 
(3) 



The values of y 2 and x in (2) and (3) substituted in the equation of 
the parabola, give 



therefore 






(4) 



140 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Equation (4) is the equation of the evolute. 

This evolute is called the semi-cubical parabola. 

Constructing the evolute, its form and position is as shown in Fig. 
25, where OA = p. 

If the origin is transferred to A, the equation becomes b 2 —■ — -a 3 , 
or as a and b are the variable coordinates, the equation may be written, 

y 2 = -^-x\ 
y 21p 

The semi-cubical parabola is so called from the nature of its equation ; the 
equation being solved for y gives the function expressed in terms of the variable 
with an exponent of three halves. 



Art. 90. A Normal to Any Involute is Tangent to its 

Evolute. 

Let P', in Fig. 24, be any point of the involute, whose coordinates 
are x' and y', and let (a, b) be the coordinates of P 1} the centre of curva- 
ture. Then the equation of the normal at P' by Art. 69 (2), is 

y-y' = -%(*-*')■ a) 

As (1) passes through (a, b), 

x'-a + ^(y'-b) = 0. (2) 

Now if P' moves along the curve, P x moves along the evolute ; hence 
a, b and y' are functions of x'. 
Differentiating (2), 

t _da dy ( ,_ h) + w_#:i =0i (3) 

dx' dx' 2KJ ) [dx'J dx' dx' K J 

But since (a, b) is on the evolute, by Art. 86 (5), 

■ [dx' 

b = y' + 



d 2 y' 
dx' 2 



EVOLUTES AND INVOLUTES. 141 

Substituting (4) in (3), 

_ da _ dy' db _r. m _ chJ_ _ db 
dx' dx' dx' dy' da 

Hence, equation (2), which, is the equation of the normal to the 
involute at (x 1 , ?/'), may be written 

y'-b=^-(x'-a), (5) 

da 

which is the equation of a tangent to the e volute at the point (a, 6). 



Art. 91. The Difference between Axy Two Radii of Curvature 

OF AX IXVOLUTE. 

The equation of the circle of curvature at (V, y 1 ) is 

( X >- a y + (y'-by = p\ (i) 

Differentiating (1), y', a, b and p being functions of x', gives 

^-.)-^-.)S+0f-»)*-<f-»)£-^ « 

By Art. 90 (2), x' - a + % (y' - b) = ; (3) 

dx 

and by Art. 90 (5), y'-b = — (x'- a). (4) 

Combining (1) and the square of (4), 

Combining (2) and (3), and the resulting equation with (4), 

-(x'-a)( da *+ 9 dhr )=e*E. (6) 

v \ da 2 J da w 

Dividing (6) by the square root of (5), and simplifying, 



Veto 2 + db 2 = t?p. 
Hence, if s represents the length of the evolute, by Art. 72, 

ds = dp ; 
therefore As = A/>, 



142 DIFFERENTIAL AND INTEGRAL CALCULUS. 

or in Fig. 24, P,P 2 = P"P 2 - P'P„ 

or the difference between any two radii of curvature of an involute is 
equal to the included arc of the evolute. 



Art. 92. Mechanical Construction of an Involute from its 

Evolute. 

From the two properties of the evolute established in Arts. 90 and 
91, the involute may be readily constructed from its evolute. Thus in 
Fig. 26, if one end of a string be fastened at N and the string be 
stretched along the curve NM having a pencil attached to the other 




Fig. 26. 

end, and then the string be gradually unwound from the evolute, 
always being in tension, the pencil will describe the involute MA. 
Every point in the string beyond N will describe an involute, as R 
describes PS. So while any curve can have but one evolute, as NM 
is the only evolute of MA, it is evident that any curve may have an 
infinite number of involutes. A series of curves having the same 
evolute are called parallel curves. 

Art. 93. Envelopes of Curves. 
If in the equation of a plane curve of the form 
f(x, y, a) = 0, 



ENVELOPES. 



143 



different values be successively assigned to a, the several equations thus 
obtained will represent distinct curves, differing from each other in 
form and position, but belonging to the same class, or family of curves. 

Now, if a is supposed to vary by infinitesimal increments, any two 
adjacent curves of the series will, in general, intersect, and the inter- 
sections are points of the envelope. 

Hence, an envelope of a series of curves is the locus of the ultimate 
intersections of the consecutive curves. 

The quantity a, which remains constant in any one curve, is called 
the variable parameter. 

R P. P 3 




Fig. 27. 



In Fig. 27, let AA', BB', etc., represent curves of a series, and a 1? 



a 2 , etc., their respective parameters ; then if a 2 



etc., 



diminish indefinitely, the ultimate intersections P lt P 2 , P 3 , etc., will be 
points of the envelope. And, at the limit, the line P l5 P 2 , joins two 
consecutive points on the envelope and on the curve BB', and hence is 
tangent to both the envelope and the curve BB', then the envelope is 
tangent to the curve BB'. 

Similarly, it may be shown that the envelope is tangent to any 
other curve of the series. 

Hence the envelope of a family of curves is tangent to each curve 
of the series. 



Art. 94. Equation of the Envelope of a Family of Curves. 
Let the equations of two curves of the series be 

/(ayy,a) = 0, (1) 

and f(x, y, a + Aa) = 0. (2) 



144 DIFFERENTIAL AND INTEGRAL CALCULUS. 

The coordinates of the point of intersection of (1) and (2) will 
satisfy both (1) and (2), and hence will also satisfy 

fix, y,a + Aa) -f(x, y, a) = 0, 

and £fe .V, a + Ag)-/(a;, y, a) = Q ^ 

Aa 

As Aa approaches 0, the limit in equation (3) is 

df(x, y, a) _ Q , 4 . 

da W 

Now the coordinates of the point of intersection of two consecutive 
curves satisfy both (4) and (1). Therefore, by eliminating a between 
(1) and (4) the resulting equation is the equation of the locus of the 
ultimate intersections, which is the required equation of the envelope. 

For example, required the envelope of a series of curves repre- 
sented by 

y = ax -til x \ (1) 

a being the variable parameter. 

Differentiating (1) with respect to a, 

*-%f = 0; (2) 

2 
hence a = — (3) 

x 



Combining (1) and (2), eliminating a, and reducing, 

y = l-- 
which is the equation of the envelope. 



2/ = l-J W 



PROBLEMS. 

1 . Find the equation of the evolute of the ellipse 

A 2 y 2 + B 2 x 2 = A 2 B\ (1) 

Here dy = _B?x ^d^ = _& 

dx A 2 y dx 2 A 2 y 3 ' 



PROBLEMS. 145 

, (A'-B 2 )^ A ( A 4 a \* 

henC e a =^ jf 2 —* and x = [j^T^J 5 

B 4 ' J \A 2 -B 2 ) 

Substituting these values of x and y in (1), 

(Aaf + (Eb)* = {A 2 - B 2 )%, 
which is the equation of the required evolute. 

2. Find the equation of the evolute of the cycloid, 



" — V2 ry — y 2 . 



sc = rvers ^ — vzry — y 



Ans. a = r vers -1 f ) + ^/—2rb — b 2 . 



3. Find the equation of the evolute to the hypocycloid, 

ajT _j_ y3 _ ^3, 

u4ns. (a + &)*+(«-&)**= '2 4*. 

4. Find the envelope of y 2 + (x — a) 2 = 16, in which a is a variable 
parameter. Ans. y == ± 4. 

5. Find the envelope of y = ax + — , a being the variable parameter. 

Qj 

Ans. y 2 = 4 m#. 

6. A straight line of given length slides down between rectangular 
axes ; required the envelope of the moving straight line. 

If c represents the length of the line and a and b the intercepts, the 
equation is 

the relation between a and b being 

a 2 + 6 2 = c 2 . (2) 

Differentiating (1) and (2) with respect to a and b, gives 

_£<to = Ji<», (3) 

and - — acZa = 6c?&. (4) 



146 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Dividing (3) by (4), 

x y x y 
x y a b a b 1 






2— ^' 



hence 



a 3 6 3 ' a 2 b 2 a 2 + b 2 c 
a = (xc 2 )*, 



and b = (yc 2 ) 3 , 

which substituted in (2), gives 

2. 2 2 

x 3 + 2/* = c 3 , 
which is the equation of the hypocycloid. 
Y 




Fig. 28. 

7. Find the envelope of a series of concentric ellipses, the area and 
direction of axes being constant. 

Ans. If c = area, the equation of the envelope is xy = ± — . 

2 7T 

8. Find the envelope of x cos a + y sin a = p, in which a is the 



variable parameter. 



Ans. x 2 + y 2 



CHAPTER XIV. 
SINGULAR POINTS. 

Art. 95. Definitions. 

A singular point is a point of a curve which has some peculiarity 
not common to other points of the curve, and not depending on the 
position of the coordinate axes. 

The most important singular points are : 

1st. Points of maximum and minimum ordinates; 2d. Points of 
inflection ; 3d. Multiple points ; 4th. Cusps ; 5th. Conjugate points ; 
6th. Stop points ; 7th. Shooting points. 

Points of maximum and minimum ordinates have been considered 
in Chapter X., and points of inflection in Art. 81. 

Art. 96. Multiple Points. 

A multiple point is a point common to two or more branches of a 
curve. 

There are two species of multiple points: 1st. Points of multi- 
ple intersection, or where two or more branches of a curve intersect ; 
2d. Points of osculation, or where two or more branches are tangent 
to each other. 

Multiple points are double, triple, etc., as two, three, or more 
branches meet at the same point. 

At a multiple point there will be as many tangents, and therefore 

as many values of -^ as there are branches. If the values of — are 
dx dx 

unequal, the multiple point will be one of the first species, but if the 

values of -^ are equal, it will be one of the second species. 
dx 

147 



148 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Let u=f(x,y) = (1) 

be the equation of the curve freed of radicals. 

Then, by Art. 47, 





du 


dy_ 


dx 


dx 


du 




dy 



And since differentiation never introduces radicals when the func- 
tion contains none, the value of -^ cannot contain radicals, and there- 

dx 

fore cannot have more than one value unless it assumes the form tt 

Hence the condition for a multiple point is -^ = - • 

F l dx 

•j »\ 

Therefore, to examine for multiple points, — and — as obtained 

dx dy 

from the equation of the curve are placed equal to zero, and the corre- 
sponding values of x and y are found. If these values of x and y are 
real and satisfy (1), they may determine multiple points. Then -^ = - 

(XX u 

is evaluated for the critical values of x and y, and every real value 
determines one branch passing through the multiple point. 



PROBLEMS. 

1. Examine the curve y 2 — (x — a) 2 x = for a multiple point. 



Here 



^ = -2(x - a)x - (x - af = 0: 
dx \ ■ J v ■■ J 



and 



du 



2y = 0. 



(1) 

(2) 



Solving (1) and (2) for x and y, gives 



x = a 


, and 


a 

x = - 

3 


y = o^ 




y = o. 



But only the first point is to be examined, as the second point does 
not satisfy the equation of the curve. 



SINGULAR POINTS. 



149 



dy _ _ —2(x — a)x— (x — a) 2 3x — a 

dx~ 2y 2Vx 

= ± Va, when x = a. 
Therefore the multiple point is a double point of the first kind, as 



shown in Fig. 29. 




Fig. 29. 



4 a; 3 + 4 axy = ; 
2 ax 2 .-3 ay 2 = 0. 



(1) 
(2) 



2. Examine the curve x* -+- 2 ax 2 y — a?/ 3 = 0, for multiple points. 

du 

dx 

du 

dy 

Combining (1) and (2) gives three pairs of values for x and y, but 
the only pair that satisfies the equation of the curve is (0, 0). 

dy = 4 X s + 4 axy = when f a; = 
cfo 3a*/ 2 -2az 2 0' l?/ = 0. 

Evaluating by Art. 59, and representing — byp and -^- byp', 

oj/ = = 12z 2 -f4a,y + 4a.rp = when <x = 
dx 6 ayp — £ax ■ 0' I y = 0, 

_ 24 a-f- 8 ap + 4 gap' _ Sap when ( as = 
6 aj) 2 + 6 ayp' — 4 a 6 ap 2 — 4 a' (2/ = 0. 

Hence p(6 ap 2 — 4 a) = 8 ap ; 

and p = ^M = 0, +V2, and -V2. 

Therefore there is a triple point of the first kind at the origin. 



150 DIFFERENTIAL AND INTEGRAL CALCULUS. 

3. Examine y 2 = a 2 x 2 — x 4 for a multiple point. 

Arts. There is a double point of the first kind at the origin where 

ax 

4. Show that the curve y 2 = x 5 + x 4 has a point of osculation at the 
origin. 

Art. 97. Cusps. 

A cusp is a point at which two branches of a curve are tangent to 
each other and terminate. 

Cusps are therefore multiple points of the second species. 

There are two kinds of cusps : 1st. When the two branches lie on 
opposite sides of the common tangent; 2d. When the two branches 
are on the same side of the common tangent. 

Since a cusp is a particular kind of multiple point, curves are 
examined for cusps as for multiple points. But as a cusp is dis- 
tinguished from a multiple point by both branches stopping at the 
point, the curve must be traced in the vicinity of the point in question 

to determine a cusp. If the two values of — \ at the cusp have con- 
dec 2 

trary signs, the cusp is of the first kind, and if they have the same 

sign, the cusp is of the second species. 

The vertex of the semi-cubical parabola is a cusp of the first kind. 
[See point A, Fig. 25.] 

The curve (y — x 2 ) 2 = x 5 has a cusp of the second species, determined 
as follows : 

Taking the square root of each member of the equation, 

y = x 2 ±x%; (1) 

hence ^L=2x±^x\ (2) 

ax 

and §= 2 ±-¥^- (3) 

ax 1 

In (1), if x = 0, then y = ; if a? is negative, y is imaginary ; if a; is 
positive, y has two real values. Hence, the curve has two branches on 



SINGULAR POINTS. 151 

the right of the T^axis which meet and terminate at the origin. The 
locus of the equation is shown in Fig. 30. 




In (2) 



dy 

dx 



0, when x = ; hence the X-axis is tangent to both 



branches, and there is a cusp at the origin. 

In (3), when a value slightly greater than is substituted for x, the 

d 2 v 
two values of — ^ are both positive ; hence the cusp is of the second 

CIX 

species. 



I. Conjugate Points. Stop Points. Shooting Points. 



A conjugate or isolated point is a point whose coordinates satisfy 
the equation of a curve, but through which the curve does not pass. 
As the conjugate point is detached from the curve, if the substitutions 
of a + b and a — b for x in the equation of the curve, b being very 
small, give imaginary values for y, then there is a conjugate point 
whose abscissa is a. 

Or, if at any point whose coordinates satisfy the equation of a 
curve, -^ is imaginary, this point will be a point through which no 

branches pass, and hence will be a conjugate point. 

For example, to examine y 2 = (x — l) 2 (x — 2) for conjugate points. 

The point (1, 0) will be such a point, for if some value a little 
greater or a little less than 1 be substituted for x in the equation, the 



152 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



resulting value of y will be imaginary, yet the point (1, 0) satisfies the 
equation. Or by the second method : 

dy _ 3 x — 5 

dx~2Vx^2 

Now the point (1, 0) which satisfies the equation of the curve 

makes — imaginary, and hence is a conjugate point. 
dx 

In Fig. 31, JOT" is the curve and P is the conjugate point. 

A stop point is a point of a curve at which a branch suddenly ends. 
For example, to examine y = x log x for a stop point. Here, for any 
positive value of x, y has one real value ; when x = 0, y = ; when x is 
negative, y is imaginary ; therefore the origin is a stop point. 




Fig. 31. 



A shooting point is a point of a curve at which two or more branches 
terminate without having a common tangent. 

1 

x 
il x 



For example, to examine y — x tan -1 ^ for shooting points. 
Here, 



4 = tar - 

dx x 



1 + x 2 



When x =* 0, then y = 0. and ^ = ± -• 
J dx 2 



SINGULAR POINTS. 153 

If x be positive and approach zero as its limit, ultimately y = and 

_^— -• but if x be negative, ultimately y = Q and-^ = — -. Hence 
dx 2 dx 2 

two branches meet at the origin, one inclined tan -1 ( - j and the other 

inclined tan _1 f — ^ Y Therefore the origin is a shooting point. 

Stop points and shooting points occur only in transcendental curves, 
and may be discovered in any curve by tracing the curve in the vicin- 
ity of the singular points. 



CHAPTER XV. 

INTEGRATION OF RATIONAL FRACTIONS. 

Art. 99. Rational Fractions. 

A rational fraction is one whose numerator and denominator are 
rational. If the degree of the numerator is equal to or greater than 
the degree of the denominator, the fraction can be reduced by division 
to the sum of several integral terms and a fraction whose numerator is 
of a lower degree than its denominator. For example, 

dx = x 2 dx + x dx - 3 dx + . 5a; + 3 dx, 



x 2 + 2x + 1 x 2 -}-2x + l 

in which the last term is the only fractional term. So it is necessary 
to consider only rational fractions in which the degree of the numera- 
tor is less than the degree of the denominator. 

A rational fraction is integrated by decomposing it into a number of 
simpler partial fractions, which can be integrated separately. 

Case 1. When the denominator can be resolved into n real and 
unequal factors of the first degree. 

Let I±J.dx represent a rational fraction, whose denominator may 

be resolved into the factors (x — a), (x— &),••• (x — I), real, unequal 

and of the first degree. 

A f(x) A ' B . C L , m 

Assume tvt= 1 7-^ ;' C 1 ) 

<f>(x) x — a x — x — c x — I 

in which A, B, C, ••• L are undetermined coefficients. Clearing (1) of 
fractions, 

f(x) = A{x — b)(x — c) — (x — l) + B(x— a)(x — c) •••(#— I) -\ 

+ L(x-a){x-b) ••• (x - k). (2) 

Performing the indicated operations in (2) and equating the coeffi- 
cients of like powers of x in the two members by the Principle of 

154 



INTEGRATION OF RATIONAL FRACTIONS. 155 

Undetermined Coefficients, will give n equations from which. A, B, C, 
etc., may be obtained. 

Or since (2) is true for all values of x, a may be substituted for x, 
which gives 

A = ^ (3) 

By substituting b for x, the value of B is obtained, and so on; 
finally when I is substituted for x, it follows that 

L = l& (4) 

These values of A, B, C, etc., are substituted in (1), dx is intro- 
duced as a factor in each term, and each term is then integrated. 



!. Find f(* + *-l 

J Q? + x 2 - 6 



PROBLEMS. 

— — -dx. 
-+- x* — 6 x 



x 3 + x 2 -6x = x(x + S)(x - 2). 

Assume f + f 1 s t4 + .*_ + _g-. (1) 

^ + x 2 -6^ a? as + 3 "a? -2 W 

Therefore x 2 + x-l = A(x + S)(x - 2) + Bx (x - 2) + Ca;(a;+ 3). 
Substituting a; = 0, gives — 1 = — 6 ^4 ; hence A = i 
Substituting # = — 3, gives 5 = 15 B ; hence -B = J. 
Substituting $c = 2, gives 5 = 10 C ; hence (7 = J. 

Substituting these values of A, B and (7 in (1), introducing dx, and 
taking the integral of each member, 

r a? + x-i dl cdx , r dx x r dx 

= i log x + J log (x + 3) + i log (a - 2) 
= log [«*(» + 3)* (a* -2)*]. 
2. J ^^lf = log [(x - l) 2 (x + 3) 3 ]. 



156 DIFFERENTIAL AND INTEGRAL CALCULUS. 

r ( x* + 2)dx ' l , x*-x-2 

/» do? = 1 1 /g±_6g> 
' J a 2 -6V 2a6 fe U-< 



5> r (2^ + 3)^ =lQg (s-1)* . 
J a* + a*-2x x i(x + 2y 

6 . J (g + 3 x - y } cfa = log ^ (2 + g) t(2 _ g)]< 



Case 2. When the denominator can be resolved into n real and 

equal factors of the first degree. 

f(x) 
Let J \ ' dx represent a rational fraction whose denominator can be 
<j>(x) 

resolved into n factors each equal to x — a. In this case the method 

of decomposition of the preceding case is not applicable. Take, for 

n 2x 2 + x 

example, - — 4— ■• 

(x — a) 3 

Forming the partial fractions as before would give 

2x 2 + x = ABC (1) 

(x — a) s x — ax — ax— a 

But if the fractions in the second member are added, 

2x 2 + x_A + B+C ,~ 

(x — ay x — a 

in which A -+- B + C must be regarded as a single constant, and evi- 
dently (2) cannot be an identical equation, as this would give three 
independent equations containing but one quantity, A + B + C, to be 
determined. 

The partial fractions are assumed as follows : 

/M^_A_ + B C + ..._L_. (1) 

tj> (x) (x — a) n (x — a)"- 1 (x - a) n ~ 2 (x - a) w 

Clearing (1) of fractions, 

f(x) = A-\-B(x-a) + C(x-a) 2 + . . . L (x - a) n ~\ (2) 



INTEGRATION OF RATIONAL FRACTIONS. 157 

The values of A, B, C, etc., in (2), are found by the Principle of 
Undetermined Coefficients, then substituted in (1), after which dx is 
introduced, and each term is integrated separately. 

PROBLEMS. 

1. Find f ^ + ^ clx. 
J (x - l) 3 

A x 2 + \ A B , C 
Assume - — ±— - = — - + — - + 



(aj - l) 3 (x - If (x - l) 2 (a; - 1) 
Hence x 2 + 1 = A + B(x - 1) + C(x - Yf 

= A+ Bx -B+ Cx 2 -2Cx+C. 
Therefore C= 1, B - 2 (7 = 0, and A - B + C= 1 ; 

whence (7=1, B=2, and A = 2. 

Therefore f^±D^= CA^ + C_A^ + f_^_ 
J (x - Vf J (x - l) 3 J (x -l) 2 J (x- 1) 

1 2 



(a; - l) 2 a; - 1 
J (z + 2) 3 (z + 2) 2 ^ gV ^ ; 



+ log (a; — 1). 



When the denominator of a rational fraction may be resolved into 
both equal and unequal factors of the first degree, the two methods 
must be combined. 

f ^ = _JL.+ log(a; + l). 

J (* + 2) 2 (a; + l) x + 2^ 8V ; 

f **fa = _ 5x + 12 in(*±*?. 

J (x + 2) 2 (a; + 4) 2 a 2 + 6 a; + 8 V» + 2 / 

J^_6a^ + 9a; 5L V yj 

J(ar>-2) 2 4(a; 2 -2)^ 8 V2 %-V2 



158 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Case 3. When some of the simple factors of the denominator are 
imaginary. 

As the denominator is real, the imaginary factors must occur in 
pairs, and of the forms 

x ± a + &V— 1, and x ± a — &V — 1, 

whose product is the real quadratic factor 

(x±a) 2 + b 2 . (1) 

For a single quadratic factor such as (1), the corresponding partial 

/\x -4- b 

fraction will have the form ^- -, because each fraction of this 

(x±a) 2 + b 2 

form increases by two the degree of the equation when it is cleared of 
fractions and therefore increases by two the number of the equations 
for determining A, B, C, etc. ; hence its numerator should add two to 
the number of these undetermined constants. 

If the denominator contains n equal quadratic factors, being of the 

form [(a? ± a) 2 + b 2 ] n , 

the partial fractions may be assumed as follows : 

f(x)_ Ax + B Cx-\- D Kx + L ,^ 

<£ 0) " [(« ± a) 2 + 6 2 ] n [(a? ± a) 2 + b 2 f~ r " (x ± a) 2 + b 2 

The values of A, B, C, etc., are determined from (2), as in the pre- 
ceding cases. 



f x 2 dx 

- J X * J_ tf _ 



PROBLEMS. 



2 



A x 2 A , B , Cx + D 

Assume = ! 

aj* + a?-2 sb+1 x-1 a? + 2 

Hence ^ = ~h B = h G =°> D = i- 

dx 



r tfdx = 1 r dx 1 rda; 2 r 
J x * + tf-<2 6J x + 1 6J x-1 3J x 

= 1 log —^^--f -77- tan l — -. 



+ 2 

^i + V?tan-^-^ 
a> + l 3 V2 



INTEGRATION OF RATIONAL FRACTIONS. 159 



„ C xdx . . , , (x 2 + l)* 

2 - I; ttt-^ — = iarc tan x + log ^ — ^— 2 - 

_ C 2xdx ,_„fx 2 -\-l\^ 

«5. 



, f ^ = /* 

J (z 2 +lVa; 2 4-^ 8 U 2 



(z 2 +l)(x 2 + 3) ° \a 2 + 3 



J (x- l) 2 (x 2 + IV 4(x-X) 2 hy } 4 



4 (sc 2 + 1) 

5 - S ^*-t = f + lQg [( * + 2)f (g ~ 2) * ] - 



da; n a 2 + # + 1 , 1 , 2x + l 



i 

— 2" 



H arc tan 



(a* + !)(*»+ a* + 1) a^ + l V3 V3 



CHAPTER XVI. 

INTEGRATION OF IRRATIONAL FUNCTIONS. 

Art. 100. Irrational Functions. 

■ Very few irrational functions are integrable. When an irrational 
function cannot be directly integrated by one of the elementary formu- 
las, an effort is made to transform it into an equivalent rational func- 
tion of another variable by making suitable substitutions. When the 
rationalization can be effected, the integral may be found. 

Art. 101. Irrational Functions containing only Monomial 

Surds. 

An irrational function containing only monomial surds may be 

rationalized by substituting for the variable a new variable with an 

exponent equal to the least common multiple of the denominators of 

the fractional exponents in the function. 

i 



/ X 2 2 X s 
— dx. 
1 + x* 



Assume x = z 6 ; then x* = z 3 , x* = z 2 , and dx = 6 z 5 dz. 

Hence 

/»/ 9*_J_1\ 

\dz 



J l + a i J 1+z 2 J 1+3 2 

= 6 Cfz G -2z 5 -z* + 2z s + z 2 -2z-l + ?£±i 
J V z 2 4-1 



( ^--2z«-%z 5 + 3z*+2z 3 -6z 2 -6z 
7 5 

+ 6 log(z 2 + 1) + 6 arc tan z 

6^! -2x-%x* +3x* +2tfi -6x* -6x* 

4- 6 log (X s + 1) + 6 arc tan x*. 
160 



INTEGRATION OF IRRATIONAL FUNCTIONS. 161 

PROBLEMS. 

1. f 2xi - l Sxi dx = ^-i^. 
J 5 a; 6 

,. r- T ^=-i + io g M±n 6 . 

J X* + X 3 iC 6 x 

3. f-^L = - 18^ + 1 + *|i + 4** + 16** + 32 log (2 - **)"} 

^ X ^ da; = 12 ($as* - f as* + V*™ ~ $ x *) 

3x* +x* 



i „ i 



+ 1908 [-J- a? T * _ 3 a;3 + 3 x? - 3£ z 6 + 81 x^ - 243 log (x™ + 3)] . 



Art. 102. Functions containing only Binomial Surds of the 

First Degree. 

A function which involves no surd except one of the form (a -f bx) v 
can be rationalized by assuming a + bx = z n , as follows : 



Let /(as, -\/a + bx) be the function. 



Assume z = V a + bx ; 

then z n = a + foe, 

nz" -1 ^ = b dx, 

nz n ~ x dz 



dx 



b 



And x = 



Therefore Cf(x, 1/a + &aT) cte = - CffiL—*, z\ z n ~ l dz, 
which is rational, and therefore can be integrated. 



162 DIFFERENTIAL AND INTEGRAL CALCULUS. 



PROBLEMS. 

xdx 



■S: 



Vl+a: 

Assume 1 + x = z 2 ; then x = z 2 — 1, and da? = 2 2 c?z. 
Hence f xdx = f2(^i) g cfa = 2 C(z 2 dz - dz) 



■Vl + x 

= -|2 3 -22 



2. 



= *(l+x) 3 *-2(l + x)k 

r dx =log yrr^-i 

J » A /1 4- cr. 



» Vl + a; VI + x + 1 

3. j (a; + Va7+2 + -fyx + 2) dx 

= |0 + 2) 2 - 2 (a + 2) 4.-|(» + 2)* + f-(i + 2)* 



4. f V1 + g;cfa / ' = a? + 4 V5 + log (v^TI). 

•^ V# — 1 

5. f ydy i = -f(4r + y)(2r-y)*. 
J(4a> + 1)* 12(4o; + l)* 

7. f d * =8 [|(i+vr^^) 1 -i(i+vr : ^) 1 ]. 

J Vl+Vl-aj 

Assume 2 = v 1 + Vl — x. 

x 2n+l dx 
Art. 103. Functions having the Form in which n is 

(a + 6a; 2 ) T 

A Positive Integer. 

Expressions of this form may be integrated as in Art. 102. 

For example, find ( — ♦ 

Assume 1 — x 2 = z 2 ; then x 2 = 1 — 2 2 , and x dx = — 2 cfc. 



INTEGRATION OF IRRATIONAL FUNCTIONS. 163 

Therefore { *'*** =_ f (1 - z*)z dz = _ f (1 _ z 2 )dz 
J Vl-a 2 J z J 



Art. 104. Functions having the Form fix. \\ c ^-^—\dx. 
In this form the assumption may be made 



that » = «+»; 

Vcx + d' 

then * = S±» 

cx + d 

and x = b - dz "; 

cz n — a 

therefore dx = n(ad-bc)^ & 

(c* n - a) 2 

The substitution of these values will make the function rational. 



PROBLEMS. 
x*dx' 3z 2 + 2 



J (1 + a?)* 3 (1 + a 2 )* 

^ (2 + 3 a 2 )* 27 (2 + 3 a?)* 



f ^ = Jlog (V3 - x 2 + 1) + | log (V^^ 2 - 3). 



a 2 + 2 V3 - x 2 



M 



+ — x dx 

5. 



+ cc (1 + xf 8 \' \1 + a? 



164 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Art. 105. Functions containing only Trinomial Surds of the 



Form Va + bx + ex 2 . 
Case 1. When c is positive. 



After factoring out Vc, the surd may be written V -4 + Bx -f- x 2 



Assume V 'A + Bx + x 2 = z — x ; 

~2 

then ^4 + ita = a; 2 — 2 z#, a; = 



jB + 2z' 



and dx ^2(z>+B Z + A)dz. 

(B + 2zf 

Therefore VZT^T^ = % - ^^ = *±B*±A u 

B + 2z 2z + B 

Thus the given function may be transformed into an equivalent 
rational function of another variable. 

Case 2. When c is negative. 

After taking out the factor Vc, the surd may be written 

VA -{-Bx- x\ 
Assume a and /3 to be roots of the equation x 2 — Bx — A = 0-, 



then Vx 2 — Bx — A = V(x — a)(x — /?), 



and -\/A -+- Bx a — x 2 = V(x — a)(fi — x). 



Let V (a; — a)(/3 — x) = (x — a) z ; 

then (x — a)(fi — x) = (x — u) 2 z 2 , 

and fe _ 2z(<*-/3)^ . 

and cto- _-J 

Therefore VA + Bx- x 2 = (x-a)z = $ ~ a ^ z - 

z- + l 

Thus the given surd is expressed in rational terms of another vari- 
able. 



INTEGRATION OF IRRATIONAL FUNCTIONS. 165 



PROBLEMS. 



1. Find 



/ 



dx 



■VA + Bx + x? 



Substituting the values of dx and V 'A -4- Bx -\- x 2 as found in Case 1, 
gives 

r dx _ r2(z 2 + Bz+A)dz x (2z + B) 

J -VA + Bx+x* * (B + 2zfx(z 2 + Bz + A) 

-/] 



B+2z a \2 



If £ = 0, (1) becomes 

/dx 
- 



=a§ 



-h a; + V-4 + 5a; + 



V A + ar 
If .4 = 1, (2) becomes 
c?a; 

vr+^T 2 

eZa; 



- = log [aj + V^4 + a 2 ]. 



2. Find 



/: 
f- 

J V^4 + 5a; - a; 2 



log [a; + Vl + a; 2 ]. 



a) 

(2) 
(3) 



Substituting the values of dx and V^4 + Bx — x 2 as found in Case 2, 
gives 

r dx = r2(a-P)zdz(z 2 + l) _ C2dz 

J VA + Bx-x 2 J (z 2 + lf((3-a)z Jl+z 2 



— 2 arc tan z = — 2 arc tan 



* x— a 



4. 






dx 



V24-3a; + a^ 

dx 
V5 x — 6 — x 2 
dx 



= log[3 + 2a; + 2V2 + 3a; + a 2 ]. 
'3 - x\i 



= 2 arc cot 



x-2 



Var* + a; 



log (£ + a; + VS 2 + <*:)• 



166 DIFFERENTIAL AND INTEGRAL CALCULUS. 






dx 1 , / xV2 

arc tan 



(l+x 2 )Vl-x 2 V2 Wl-ar 

da; 1 



Va 2 + 6 V 2 & 



log (6a; + Va 2 + fr 2 a; 2 ). 



dx 1 t / Va 2 + 6 2 a; 2 — a 



log 



«Va 2 + 6 2 a; 2 « V te 



or _l lo ^f Va 2 + 6 2 a; 2 H-a 



Art. 106. Binomial Differentials. 

Binomial differentials have the form x m (a + bx n ) p dx, in which m, n 
and p are any numbers, positive or negative, integral or fractional. 
1st. If m and n are fractional, and the differential has the form 

s l 

x t (a + bx r ) p dx, 
x = z rt may be substituted, and the expression becomes 
2 rs (a -f bz u ) p rtz rt - l dz = rtz rs+rt ~\a + bz lt ydz. 
2d. If w is negative, and the differential has the form 
x m (a + bx n ) p dx, 
x = - may be substituted, and the expression becomes 
x m (a + bx~ n ) p dx = — z- m ~ 2 (a + bz n ) p dz. 

Hence, any binomial differential may be transformed into another, 
having integers for the exponents of the variable, and having a positive 
exponent for the variable within the parenthesis. 

In the following articles every binomial differential is assumed to 
have this reduced form. 

Art. 107. Conditions of Integrability of Binomial 
Differentials. 



As the exponent of the parenthesis is any number, let it be repre- 

ray be written 

p 
x m (a + bx n ) q dx. (1) 



sented by -, and the form may be written 



INTEGRATION OF IRRATIONAL FUNCTIONS. 167 

1st. Assume (a + bx n ) = z q ; 

p 
then (a + to")* = 2f», (2) 



•=('-?)"• 




--(*?£ 






1 



(3) 



and clx=y-z*- 1 (— Y dfe. (4) 

?i& \ 6 / 

Substituting the values from (2), (3) and (4) in (1), 

OT+l 

a" 1 (a + Mx = -i ap+i-if^lziA " (fo, (5) 

w& \ b J 

which is rational when — ^— is an integer or 0. 



2d. Assume ax~ n + b = z q ; (1) 

then x = a" (z 9 — 6) ", 

a;* = a (z 9 - 6)^, (2) 

1 -i-i 

and cfa = -2eWz 9 -&) " z 9 " 1 ^. (4) 

Multiplying (2) by 6, adding a, and taking the ? power, 

(a + &af) 9 =a 9 (z 9 — &) 9 z p . (5) 

Taking the product of (3), (4) and (5), gives 

? -+P.1-! / »t+l y .\ 

af (a + to 71 ) 9 die = — ? a n 9 n (z 9 — 6) V n 9 ; z^ 9 " 1 ^, 

which is rational when m ~^~ + - is an integer or 0. Therefore, the 

n q 

binomial differential can be rationalized : 



168 DIFFERENTIAL AND INTEGRAL CALCULUS. 

1st. When the exponent of the variable without the parenthesis, in- 
creased by one, is exactly divisible by the exponent of the variable within 
the parenthesis. 

2d. When this fraction increased by the exponent of the parenthesis is 
an integer. 



PROBLEMS. 



1. YmdCx 5 (2-\-3x 2 )Ux. 



Here !?L±!_l = *±1-1 = 2 



therefore the first condition of integrability is satisfied. 

Hence, let (2 + 3x 2 )=z 2 -, 

then (2 + 3x 2 )i=z, 

and dx = 



3 /' 

zdz 



Therefore Cx 5 (2 + 3 x 2 ) * dx = Cf 



3 

z 2 — 2\f zdz 



= if(z«-4z 4 + 4z sl )<fe 

__ 1 /z' 4z" 4z : 
_ 27\7 5 3 



= ^[1(2 + 3^-4(2 + 3^' 
+ 4(2 + 3^)*]. 

J 3 xr 

3. f^(«+^ 2 ) f ^ = (a + 6<)^ 



35 6 s 






x 2 (l + x 2 ) 



INTEGRATION OF IRRATIONAL FUNCTIONS. 169 

5. I x~ 2 (a + a 3 ) 3 dx = — -■ 

^ 2a 2 rc(a + <c 3 )* 

7. fa 5 (a + x 2 ^ dx = ^ (a + x 2 ) 1 ^ — f- (a + £ 2 ) ^ a + | (a -f a 2 )* a 2 # 



PRACTICAL PROBLEM. 

A vessel in the form of a right circular cone is filled with water and 
placed with its axis vertical and vertex down. If the height = h, and 
radius of the base = r, how long will it require to empty itself through 
an orifice in the vertex of the area a ? 

Neglecting the resistances, if the vessel is kept always full, the 
velocity of discharge through an orifice in the bottom is that due to a 
body falling from a height equal to the depth of the water. If v 
denotes the velocity and x the depth of the water, 

v = V2 gx. 

If dQ denotes the quantity discharged in the time dt through an 
orifice of the area a, 

dQ =adt^/2gx. 

But in the time dt the surface whose area is S has descended the 

distance dx, thus 

dQ = Sdx. 

Hence Sdx = adt-y/2 gx, 

,. Sdx 
or dt = 

a^/2gx 
At the distance x from the vertex, 



Therefore 



-x 



h irr^aPdx _ 2 Trr 2 Vfe 
ah 2 V2 gx 5 a V2 g 



CHAPTER XVII. 

INTEGRATION BY PARTS, AND BY SUCCESSIVE REDUCTION. 

Art. 108. Integration by Parts. 

Integrating both members of d (uv) = udv -\-v du, and transposing, 
J udv = uv — I v du. (1) 

Equation (1) is the formula for integration by parts. 
By this formula, I u dv is made to depend on I ^ du, and this new 
integral is frequently much simpler than the given one. 

PROBLEMS. 
1. Find I x n logic dx. 
Let u = log x, and dv = x n dx ; 

then du = — , and v = 



x ?i + l 

Substituting these values in the formula for integration by parts, 

x n log xdx = ^ I x n dx 

n + 1 n -f 1 J 

x n+1 A 1 

iogx 



n + 1\ n + 1 

2. I logic dx = x (logic — 1). 

3. Co sin dO = — 9 cos + sin 0. 



170 



INTEGRATION BY PARTS. 171 

/"log (log x) dx t i /i x i 

4. I — ^ — ^—^ — = log a: • log (log a*) — log a;. 

%J X 

5 . Cxe ax dx — e ax [ _ _ i\ 
J \a a 2 J 

6. I arc since da? = x arc sin a; + (1 — x 2 ) *. 

7. J xcosxdx= #sina,' + cos#. 

/x 2 
x tan 2 a; cto = x tan # h log cos x. 

rlogxdx x , t x . 1N 

10. f^^ = ^- 3 -^ + 6|-6V". 

J \ a a 2 a 3 J a 

Art. 109. Formulas of Reduction. 

When the integral, j x m (a + bx n ) p dx, 

satisfies either of the conditions of integrability as given in Art. 107, it 
may be rationalized as explained in that article and then integrated. 

But by means of certain formulas of reduction, derived by the aid 
of the formula for integration by parts, the given expression may be 
made to depend upon simpler integrals of the same form. This method 
is called integration by successive reduction, and the integrals given by 
this method are generally in convenient form for integration between 
limits. 

1. Formula A. 

Assume j x m (a 4- bx n ) p dx = I udv = uv — I v du. (1) 

Let dv = x n ~\a -f bx n ) p dx, then u = x m ~ n+ \ 

Hence v = ( a + ox n ) p+1 and du=(m — n + l) x m ~ n dx. 



172 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Substituting in (1), and putting a + bx n = X, 

Cx m X p dx = x m ~" +l X p+l _ m-n + l f x m-n X p+i dXt m,) 

J nb(p + l) nb(p+A)J v J 

Now C x m - n X p+1 dx = C x m - n X p (a+ bx n )dx 

= a \x m ~ n X p dx + b Cx m X p dx. (3) 

Substituting in the right member of (2) the value from (3), 

/ x m ~ n+1 X p+1 Cm — n + l)a r TO _ n Vv , 
x m X p dx = * ■ — ^— I x m n X p dx 
nb(p + l) nb(p + l) J 

_ m-n + l r xmXPdx ^ 
n(p + l)J 

Transposing the last term to the first member and solving for J x m X p dx, 

fa-XWte = s""'*'* 1 - «(™-" + l) fa-X'dx. (A) 

J b(np+ m + 1) b (np + m + l)J 

By Formula (A), the given integral is made to depend upon another 
of a similar form, having the exponent of x without the parenthesis 
diminished by the exponent of x within. 

2. Formula B. 

Cx m X p dx = fx m X p ~ 1 (a + bx n )dx 

= a Cx m X p ~ l dx + b Cx m+n X p ~ 1 dx. (1) 

Applying Formula (A) to the last term of (1), by substituting m + n 
for n, and p — 1 for p, 

J np + m + 1 np + m + 1*/ 

which substituted in (1), by uniting similar terms, gives 

Cx m X p dx == xOT+1 ' YP ^ + ^ — - (Vx^cte. (B) 

J np + m + 1 nj> + m + 1*/ 



INTEGRATION BY SUCCESSIVE REDUCTION. 173 

By Formula (B), the given integral is made to depend upon another 
of a similar form, having the exponent of the parenthesis diminished 
by unity. 

Formulas (A) and (B) fail when np + m + 1 = 0, but in this case 

TljZ — |-p = 0; hence the method of integration of Art. 107 is appli- 

n 
cable. 

3. Formula G. 

Solving Formula (A) for I x m ~ n X p dx, gives 

(V-x-ca- = f-+'*' + ' _ & (;* + »» + !) fa XHlx ; 

J a (m — n -+- 1) a (m — «. -j- 1) J 

Substituting — m for m — ?i, 

Pb-2^ = X ' m ] lX ' +l + &(m-n-„p-l) fV— ^^ (C) 

J — a (m — 1) — a (m — 1) */ 

By Formula (0), the given integral is made to depend upon another 
of a similar form, having the exponent of x without the parenthesis 
increased by the exponent of x within. 

Formula (0) fails when m — 1 = ; in this case 

m = 1, and — m + 1 = ; 

hence the method of integration of Art. 107 is applicable. 

4. Formula D. 

Solving Formula (B) for I x m X p ~hlx, gives 

f *-X>-d* = - £*-' + ^ + TO + 1 > ffXM* 

Substituting — p forp — 1, 

f«-nfa = *"' x ^' - ft» + "-"P + P f>x-*+'<fo. (D) 

«y an(p-l) an(p-l) J 

By Formula (D), the given integral is made to depend upon another 
of a similar form, having the exponent of the parenthesis increased by 
unity. 

Formula (D) fails when p — l = } but in this case the integral 
reduces to a fundamental form. 



174 DIFFERENTIAL AND INTEGRAL CALCULUS. 

PROBLEMS. 

1. Find f x " dx . 
J Va 2 — x 2 

f f dx = f x 3 (a 2 - x 2 )'hx. 
J Va 2 — x 2 J 

Here m = 3, ?i = 2, p = — J, a = a 2 , and 6 = — 1. 
Substituting these values in {A), 

C^( 2 ^-to a 2 (a 2 -x 2 )* 2a 2 f /2 2N -i, 
I or (a 2 — or) 2 cto = — * — - — £ — I x(a? — x 2 ) 2 dx 

x 2 (a 2 — x*y . 2a 2 r , 2 2N i-. 
= ^-3 — ^+-3- [-(a 2 -«■)'] 

= _ \(a 2 -x 2 )*(x 2 + 2a 2 ). 

2 - Find /(T&) 3 - 

Here m = 0, n = 2, — p = — 3, a = 1, and 6 = 1. 

Substituting these values in (D), 

J"(l + s 2 )" 3 ^ = ^ (1 + ^" 2 + 1 j"(l + x 2 )~ 2 dx. (1) 

Applying (D) to the last term of (1), making m=0, n = 2, — j9 = — 2, 
a = 1, and 6 = 1, 

f (1 + o; 2 )- 2 ^ = ft^ + a 2 )" 1 + 1. f(l + o; 2 )" 1 ^ 

= 2aT^) + * arctana; - 

Therefore f- dx - = ,* =-. + — ~ , +|arctana;. 

J (1 + x 2 ) 3 4(1 + x 2 ) 2 8 (1+x 2 ) 8 

3. Find f 



a^Va' — x 2 

r dx = ^-3^2 _ tfyl dXt 

J x s -\/a 2 — x 2 J 



PROBLEMS. 175 

Here — m = — 3, n = 2, p = — %, a = a 2 , and b = — 1. 
Substituting these values in (C), 

fx-\a? - x*yhx = - x - 2 (^- x2 y + _i_ JV-i (a 2 _ j^-1^ 

By Art. 105, Ex. 8, 

a — Va 



JV' (a 2 - «F)-*«fc = I log ( °~ V ° H 



mi n r dx _ _ Va 2 — x 2 _1_ , _ fa — Va 2 — x 

J or 3 Va 2 - x 2 ~ 2 a¥ 2 a 3 



arc sin -• 
a 



_ /* a^cfo # /-« -« , a 2 . x 

5. I — =--yy — or -j — arc sin — 
J Va 2 - x 2 2 2 a 

6. f (1 - a 2 ) 1 ^ = \ x(l - x 2 f + f x(l '.- x 2 y + | arc sin x. 

7. C^(a 2 + * 2 )dx = - Va 2 + x 2 + - log (x + Va 2 + x 2 ). 

J vr^ 2 V5 5.3^5.3; 

9. fz 2 (l - z 2 )*dz = i a; (2 a 2 - 1)(1 - xrf + 

10. far 5 (1 + x 2 ) kx = ( 5 f ~ 2) (1 + a?)* 
»/ 3 • 5 

u. f <fa = _vr+^ u / vr+i-i 

JarVl + a * Wl+'as + l 

J (a + 



arc sin #. 



„. ,_^_ r= f i_ o+ 2 



& ^)f \a + 6a; 2 ay 3 a Va + bx 2 



l3 . f aftte = -^4-^ + ^-V2^^^ + fa 3 arcvers^ 
J V2aa;-ar 2 l3 T 6 T 2J ^ 2 a 

, Eemark. Reduce f Mx ■ to the form Cx*{2 a- xy^dx 



176 DIFFERENTIAL AND INTEGRAL CALCULUS. 

C x 2 dx x + 3 a /o —o , , o x 

14. I — ' = ■ V^ ax — xr-\-\a? arc vers — 



/' x 4 dx x(3 — x 2 ) 
= — ^— '-? — f arc sm a;. 
(l-x 2 f 2(l-x 2 y 

i6. f = =- ±\(3x« + 4^ 3 + 8) vr^^. 

•^ Vl — x 3 



CHAPTER XVIII. 

INTEGRATION OF TRANSCENDENTAL FUNCTIONS. INTEGRATION 

BY SERIES. 

Art. 110. Introduction. 

The method of integration by parts gives important reduction 
formulas for transcendental functions. Only a comparatively small 
number of logarithmic and exponential functions can be integrated by 
general methods. It is frequently necessary to resort to methods of 
approximation. Some of the principal integrable forms will be given 
in this chapter. 

Art. 111. Integration of the Form j f(x) (log x) n dx. 

It is assumed in this form that f(x) is an algebraic function and n 
is a positive integer. 

Let f(x)dx = dv, and (log x) n = u ; 

/dx 
f(x)dx = v, and n(log#) n-1 — = du. 
x 

Substituting these values in j u dv = uv — I v $u, 
jfix) (log xydx = (log x)«ff(x) dx - j[n (log a?)- 1 ~ff(^) dx] ; 
or, by making j f(x) dx = X, 

Cf(x) (log x) n dx = X (log x) n — n f — (log x^dx. (1) 

Hence, whenever it is possible to integrate the factor f(x)dx, the 
given integral will depend upon another of a similar form, in which 
the exponent of the logarithm is diminished by unity. By repeated 
n 177 



178 DIFFERENTIAL AND INTEGRAL CALCULUS. 

applications of this formula the given integral will depend finally on 
the algebraic form I (f>(x)dx. 

PROBLEMS. 
1. Find Cx(\ogx) 2 dx. 

Let x dx = dv, and (log x) 2 = u ; 

x dx 

then — = v, and 2 log x — = du. 

2 x 



dx 
x 



Hence J a? (log x) 2 dx = — ■ (log x) 2 — I x 2 log x 
Similarly, f x log xdx = — (log x) — I • 

/X 2 X 2 X 2 
x (log x) 2 dx = — (log a-) 2 — — log x -\ 

2 . I x A (log x) 2 dx = — (log 2 x — -| log a? + ^)* 

^ a;2 3Va; 3 

Jct ,n+1 
x' n (log x) 2 dx = 
n + 1 



2 2 

(log x) 2 — log x ' 



71 + 1 (^ + 1) 2 _ 

Art. 112. Integration of the Form I x m a nx dx. 

In this form it is assumed that m is a positive integer. 
Let a/^dx = dv, and sc m = u ; 



then — = v, and mx m ~ 1 dx = du. 

nloga 

Therefore fx m a nx dx = ^^ -^— faJ-^crtte. 

*/ nloga ?iloga«/ 



INTEGRATION OF TRANSCENDENTAL FUNCTIONS. 179 

By successive applications of this formula, the exponent of x can 
be finally reduced to zero, and the given integral made to depend on 



the known for 



?m, J a nx dx. 



PROBLEMS. 

,x dx. 
Let e ax dx = dv. and x 2 = u 



then fx 2 e ax dx = - e^x 2 — - ( xe ax dx. 

J a aJ 

Similarly, ( xe ax dx = - e ax x I e ax dx. 

J a aJ 



2 -? + \ 

a \ a a- 



Therefore (x?e ax dx = —(x 2 — ~ + - 2 )• 



2. I xa x dx =J*L-(x 



, fa 

J log a \ log a, 

3. Cx 2 e x dx =e x (x 2 -2x + 2). 

J a\ a cr a 3 J 

5. C&b =_ c --(aj» + 2» + 2). 



Art. 113. Integration of the Form ( sin" 1 cos n 6 dd. 



/■ 



1st. When either m or ?i, or both, are odd positive integers. 
In this case the integration can be effected as in the following 
example : 

f sin 3 6 cos 4 dd = f (1 - cos 2 0) cos 4 6 sin dO 



= — j (cos 4 6 — cos 6 0) d cos 



cos 5 6 cos 7 

~5~ 7 



180 DIFFERENTIAL AND INTEGRAL CALCULUS. 

2d. When m + n is an even negative integer. 

In this case the integration can be effected as in the following 
example : 

f sin 2 cos" 6 6 d0 = ftan 2 cos~ 4 dO 
= ftan 2 sec 4 6 dO 
= ftan 2 6 (1 -f tan 2 6) • d tan 6 

= T(tan 2 (9 • ot tan (9 + tan 4 . d tan 0) 

= tan 3 6> tan 5 (9 
3 + 5 " 

3d. When the form is not immediately integrable, neither of the 
aforesaid conditions being fulfilled. 

In this case the integral can be obtained by successive reductions. 

n 

Let sin 6 = x ; then sin™ 6 = x m , cos n $ = (1 — x 2 ) 2 , 

and dd = (1 - x 2 )'%dx. 

Hence fsin w cos" c?0 = Cx m (1 - a 2 )" 1 "^. (1) 

Thus the given trigonometric form may be transformed into a 
binomial differential which may be integrated by means of the formulas 
of reduction. 

For example, to find j sin 4 6 cos 4 dO. 

Let sin = x-, then sin 4 6 — x 4 , cos 4 == (1 — a 2 ) 2 , and 

d$ = (1 - a 2 ) _ *cfa. 
Hence f sin 4 (9 cos 4 6> c^6> = fa 4 (1 - a 2 )^fo. 

Applying Formula (A) twice, 
(V (1 - x 2 )Hlx = - g 3 ^-^ 2 ) 2 _ | . i ^(i _ ^f + 3 . i T(i _ afyi'dx. (2) 



INTEGRATION OF TRANSCENDENTAL FUNCTIONS. 181 

Applying Formula (B) twice to the last term of (2), 

f (1 - xrfdx = x ( 1 ~ x2 ) 2 + |.i a; (l_ fl j2)^ + |. ij(i _ ^yi dx . 

Hence 

^(l-^^ = - ^ (1 -^ -f.i^(l -^+3.1.1 ^(1,^1 

+ 1 * * * I "• i x C 1 - «rf + 1 • i * I • i arc sin *■ 

Therefore 

fsin 4 cos 4 6d6 = — Q -^~- (sin 3 + i sin 0) + ^ sin (cos 3 + f cos 0) 



+ T 1 



2 8 v ' 

PROBLEMS. 



1 . fsin 5 cos 5 dO = - JL. cos 6 (sin 4 -f- -J sin 2 + £). 

2 - f-^T-= tan0 + -|tan 3 + ^tan 5 0. 
J cos 6 8 6 

3. C sm5 ° de = sec + 2 cos - 4 cos 3 0. 
J cos 2 3 

4 f__^_ = 1 4cos0 8cos0 

J sin 4 0cos 2 0~cos0sin 3 3 sin 3 8 sin 

5 . fsin 3 cos 3 c?0 = l sin 4 - i sin 6 0. 

6 r sin 2 d(9 ^ tan 5 tan 3 
J cos 6 5 3' 

7. fsin 4 0c70 = - J cos (sin 3 + f sin 0) + f 0. 

8 . f cos 4 d6 = i sin (cos 3 + § cos 0) + f 0. 

n r 0/1 • 4 a 7/, sin cos /sin 4 sin 2 1\ . 

9. I cos- sin 4 (70 = [ — )-\ 

J 2 V 3 12 SJ 16 

10. f . — - = sec + log tan -. 
J sm0cos 2 & 2 

11. f sec 3 (70 = sec ° tan ^ + 1 log (sec + tan 0). 

10 C de cos , , , , 

12. I = \- i log tan — 

Jsin 3 2sin 2 2 & 2 



182 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Art. 114. Integration of the Forms I x n cos (ax) dx and 
ft x n sin (ax) dx. 

The formula for integration by parts is used, assuming that u == x n . 
Evidently, each application of the formula will diminish the exponent 
n by one ; therefore, when n is a positive integer, the given form may 
be made to depend finally on J sin (ax) dx or J cos (ax) dx, each being 
a simple known form. 



For example, to find j x 2 sin x dx. 



Assume u = x 2 , v and dv = sin x dx ; 

then du = 2 x dx, and v = — cos x. 



Hence j x 2 sin xdx = — x 2 cos x + 2 I x 
Similarly, I x cos x dx = a; sin x — I sin a? 



cos x dec. 
= x sin a? + cos x. 



Therefore j x 2 sin xdx = — x 2 cos x -\- 2x sin x -f- 2 cos x. 

Art. 115. Integration of the Forms I e ax sin n xdx and j e^cos^xdx. 

Assume u = sin n x, and dv = e ax dx ; 

e ax 
then dw = n sin n_1 x cos xdx, and v = — • 

a 

Hence I e ax sin" x dx = - e a2 sin n x j € ax sin n_1 x cos xdx. (1) 

./ a aJ 

Again, assume u = sin n_1 x cos x, and dv = e aI dx ; 

then du = (n — 1) sin" -2 x cos 2 x dx — sin n x dx 

= (n — 1) sin n_2 xdx — n sin n xdx, 

e ax 
and -y = — 

a 



INTEGRATION OF TRANSCENDENTAL FUNCTIONS. 183 

Hence 

xdx 



/ e ax gi n n-l x CO g x d x — _ Q ax s i n n-l x C0S X ( e ax SHl"' 
a a J 

+ - | e as sin n x dx. 
aJ 

Substituting in (1), and solving for J e ax sin n x dx, gives 

/' ' : . " _ 7 e ax sin n_1 x (a sin x — n cos a?) 
e ax gln n /g ^ _ ^ £ 
rr + cr 

+ W ^ ~ V fe ax sin n - 2 cc dte. (2) 

n 2 + a 2 J 

Each application of this formula diminishes the exponent of sin x 
by 2. By repeated applications n can be reduced to or 1, and the 
given integral will finally depend on 

j e ax dx = — , or j e ax sin x dx. 
The value of j e ax sin x dx is obtained directly from (2) by making 



n = l. 

In like manner 1 e ax cos n x dx can be obtained 



PROBLEMS. 



1. I x 2 cos xdx = x 2 sin x + 2x cos £ — 2 sin x. 

2. I # 3 sin axte = — ar 3 cos a? -f 3 a; 2 sin # + 6 a; cos a? — 6 sin x. 

/e ax 
e ax sin x dx = (a sin x — cos a;). 
a 2 + 1 

4. j e x sin 3 oj da? = — (sin 3 x -f 3 cos 3 a; + 3 sin x — 6 cos #). 

r ai o 7 e a2 cos x (a cos a; + 2 sin a;) . 2 e ax 

5. I e ai cos- x dx = * -J z -f- . 

J 4 + a 2 4 + a 2 a 



184 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Art. 116. Integration of the Forms 
I f(x) arc sin x dx, j f(x) arc cos x dx, I f(x) arc tan x dx, etc. 

In these forms, f(x) is an algebraic function. 

Any one of these forms may be integrated by using the formula for 
integration by parts, assuming dv=f(x)dx. For example, to find 



/• 



# 2 arc sin x dx. 



Assume dv = x^dx, and u = arc sin x ; 

x dx 

then v — — ? and d% 



Vl-a 2 



Hence ( a; 2 arc sin xdx = — arc sin x — \ \ - 

J 3 v vr^ 2 

Substituting a = 1, in Ex. 1, Art. 109, 

C xhlx = - f(l - a 2 )^ 2 + 2). 

•^ Vl — A3 2 

x 2 arc sin cccfcc = — arc sin x -\- i (1 — x 2 ) 2 (a? 2 + 2). 

PROBLEMS. 

1 . I arc sin xdx = x arc sin x + (1 — ic 2 ) ¥ . 

2. j — ^ — n 2 ^ = a; arc tan x — i (arc tan a?) 2 — i log (1 -+- x 2 ). 

%J _L -J— X 

-. ix arc sin xax 1 / 9 , o\ /h ~ 9 • . x , <> 

3. I == — = — %{ar -+- 2) VI — x 2 arc sin a; -f 77 + i x - 

J Vl — a 2 9 

4. j a; arc cos xdx = \x 2 arc cos x — \x(l — x 2 y -f- J arc sin cc. 

- 
a 



r dO _ r 
J a + b cos ~~J 



+ 6 cos 
dO 



{ oos2 © + - 2 ©]+{ 00 <I)- sin2 (f)] 



INTEGRATION OF TRANSCENDENTAL FUNCTIONS. 185 

dO 






(a + b) cos 2 ^ + (a - b) sin 2 Q 



sec 2 ? W# 



(a + 6) + (a — 6) tan 2 



_,/. 



cZ tan f - 



(a + 6) + (a- 6) tan 2 f|\ 



VaM^ 



arc tan 



tan , 

a + b 



when a>b. (1) 



If a < 6, then 



J a+b cos J i 



cl tan ( - 



(b + a)-(b - a)taii 2 (Pi 

., Vb +a + Vfr — atanf- ) 



W- a 2 



loi 



V& + a — V& — a tan f - 
by Ex. 4, Art. 99. V2 

/dx 
; — may be obtained. 
a + b sm x ' 



(2) 



PROBLEMS. 



1. Find 



/j 



d» 



2 + cos 
Substituting a = 2 and b = 1 in (1) gives 



JV 



d<9 



+ cos V3 



— arc tan ( V? tan - V 



d6 



• fr 



/: 



3 + 5 cos 
dO 



tan- + 2 
2 

tan^-2 



5 — 4 cos 2 



— - = i arc tan (3 tan x). 



186 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Art. 118, Integration by Series. 

When a given function cannot be integrated by any of the preced- 
ing methods, or when the integral obtained is too complicated in form 
for convenient use, recourse is had to the method of approximation 
called integration by series. By this method the function is developed 
into a series whose terms are integrated separately. If the resultant 
series is convergent, an approximate value of the integral is found by 
summing a finite number of terms. 

PROBLEMS. 
1, Find Cx 2 (l-x 2 )^dx. 

Expanding (1 — x 2 )* by the Binomial Theorem, 

(l - x 2 y = i-i» 2 -ix 4 - T \x 6 .... 

Therefore Cx 2 (1 - x 2 f dx = Cx 2 (1 - \ x 2 - \ x* - ^ x G • • •) dx 



X s _ X 5 _ X 7 __ X s 

3~~10 56~144 



■■m 



dx _ X 2 X*. X* X 5 



3. Cxi(l -x*)\dx=% x l-\x%--£ i x*-Tfa s x 1 £ — 



f 



dx x 6 3 a; 5 3- 5 a; 7 



VIT^ 2 2 ' 3 2 ' 4 ' 5 2-4-6-7 



e x dx. 



Vl - x 2 

(1 - e 2 x 2 f = l-ie¥ + — eV - -1^- e 6 a; 6 ••• 
Therefore 

Vl - e 2 x 2 , /"A i „2„ 2 , 1 „4„ 4 1-3 « « \ da; 



f V l— Jaf fa = r/ 1 _i e 2 a . 2 + _l_ eV _ Ij^eV- 

J vr^a?" J v 2 * 4 2 * 4 - 6 



Vl-aj* 



arc sin a; -f i e 2 (-*• x Vl — x 2 — \ arc sin a?) 

1-3 



6 . C af* (x — 1)$ ckc = f a; 6 — 4 a;^ + T \ 



arc sin a; 
4 



5 11 

8 
29T 



£C 6 _J_ __§_ ^ "6 . , 



CHAPTER XIX. 

INTEGRATION AS A SUMMATION. AREAS AND LENGTHS OF 
PLANE CURVES. 



Art. 119. Integration as a Summation. 

The integration of a function may be regarded as the summation of 
a certain infinite series of infinitely small terms. The problem of 
finding the areas of plane curves furnishes a good illustration.* 

For example, to find the area AP x P n N, included between the curve 
BS, the X-axis, and the ordinates AP ± and JSfP n . 



P»-i£ 




Fig. 32. 

Let y =f(x) be the equation of the curve. And let OA = a, ON= b, 
and divide AN into n equal parts each denoted by Ax, and erect ordi- 
nates at the points of division. 

Then, area of rectangle PJ$ =f(a)Ax > 

area of rectangle P 2 C =f(a + Ax) Ax, 

* Newton's Lemma II., Principia, Lib. I., § 1. 
187 



188 DIFFERENTIAL AND INTEGRAL CALCULUS, 

area of rectangle P 3 D =/(a + 2 Ax) Ax, 

and area of rectangle P n -\N=f{b — Ax) Ax. 
Therefore, the sum of the ?i rectangles is 
f(a)Ax +f(a + Ax) Ax +/(a + 2 Ax) Ax -\ \-f(b - Ax) Ax, (1) 

which may be represented by N fix) Ax, in which /(a;) Ax represents 

each term of the series, a; taking in succession the different values 
between a and b. 

Now as Ax approaches zero, n increases indefinitely, and the limit 
of the sum of the rectangles is the required area AP-yP n N. 

When Ax becomes dx, the symbol N is replaced by I , and the 

expression for the area, which is the sum of an infinite number of 
infinitely small rectangles, becomes 

f f(x) dx = f (a) dx +/(a -f dx) dx + f(a +'2 dx) dx • • • + f(b - dx) dx. (2) 

Assume J f(x) dx = cf> (x) ; 

then / (x) dx = d(/> (x) = <£ (x + dx) — <£ (x). (3) 

Substituting successively for x in (3) the values, 
a, a -f- dx, a + 2 dx, • • • 6 — dx, 
gives / (a) dx = <j> (a + dx) — <f>(a), 

f (a + dx) dx = <£ (a + 2 dx) — <£ (a + dx), 
/(a + 2 dx) dx = <£ (a + 3 dx) — <£ (a + 2 dx), 

fib — dx) dx — $ (b) — cfi (b — dx). 
Adding these equations, 

f(a)dx +f(a + dx)dx -f- •••/(& — dx)dx = <f>(b) — <f>(a), 

or I /(a?) dx = <j>(b)— <j> (a). 

Therefore the area is found by integrating f(x)dx, substituting b 
and a successively in the integral, and subtracting the latter result 
from the former. 



INTEGRATION AS A SUMMATION. 



189 



PROBLEMS. 

1. Find the area of the circle x 2 + y 2 = ?* 2 . 

Area of a quadrant = I f(x)dx = J ydx= J {r 2 — x 2 ydx. 
By Ex. 4, Art. 109, 



X' 



(r 2 — x*)Hlx = 



x(r 2 — x 2 )* r 



+ — arc sm - 



therefore the area of the whole circle is irr 2 . 

In order to obtain the area of the 
semi-segment OABC, Fig. 33, the 
superior limit of integration will be 
OA = x, and the inferior limit will 
be zero. 

Therefore 

area OABC = C\r 2 - xrfdx 



_ xir 2 — x 2 ) 2 7 



-f- — arc sin-. 
2 r 




Fig. 33. 



Evidently 



and 



l -^ —L- = area of triangle 0.4.B, 

—arc sin - = area of sector 05(7. 
2 r 



2. Find the area between the curve y 2 = 4sc, the axis of X, and the 



ordinate through the focus. 

3. Find the area of the ellipse ahf + b 2 a? = a 2 b 2 . 



Ans. A = %. 



Ans. -wab. 



4. Find the area of the hyperbola xy = 1 between the limits x = a 
and x = 1. Ans. Log a. 

In this example it will be seen that the area of the hyperbola is the Naperian 
logarithm of the superior limit. For this reason Naperian logarithms are also 
called hyperbolic logarithms. 



190 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



5. Find the area of the cycloid x = rare vers - — \/2 ry — y 2 . 



Here 



Therefore 



dx 



y dy 



V'2ry-y 2 

c/0 



= #7rr 2 . 



V2 ry — y 2 

6. To find the area of y (1 + x 2 ) = 1 — x 3 , between the curve and 
the axes, in the first quadrant. 

The limits will be found to be x — 1 and x = 0. 



Therefore 



'-JTt 



+ x* 



dx 



dx 



r x 2 "i 1 

= — — + \ log (x 2 + 1) + arc tan x 

= .631972. 
7. Find the area included between y 2 = 2px and x 2 = 2py. 

Y 




Fig. 34. 



The two parabolas intersect at (0, 0) and (2p y 2p) ; hence the limits 
of integration are 2p and 0. 

Area OBPA= f s/2pxdx. 



Area OCPA 



= r 2 p <tf_ 

~Jo 2 7) 



2p 



dx. 



INTEGRATION AS A SUMMATION. 191 

Therefore area OBPC = C*f-y/2px - — ^ dx = 4p?. 

8. Find the area included between if = 2 x and y 2 = 4 x — x 2 . 

Ans. 0.475. 

9. Find the entire area within the hypocycloid x* -f y* = a J . 

A 3tt(T 

Ans. • 

8 

10. What is the area of a theoretical indicator diagram when the 
steam is cut off at half -stroke, if the law of expansion is pv = 1? 

Ans. 1 + log 2. 

Art. 120. Areas of Plane Curves in Polar Coordinates. 

Referring to Fig. 35, it is required to find the area POP n , included 
between any plane curve AB and two vectors OP and OP n . 

B \ 
P. 




Fig. 35. 



Let the vectorial angles POX and P n OX be denoted respectively 
by (3 and a. 

If the coordinates of any point P be (r, 6), then the coordinates of 
P x will be (r + Ar, 6 + A0). 



The area of sector POS = \ r ■ rA0 = \ i^AO. 



Then the sum of all the sectors POS, PiOS 1} etc., may be repre- 
sented by / A r 2 A0 ; and as A0 approaches zero, the limit of the sum 

of the sectors is the required area POP n , which will be given by the 
expression 



A = I CrhW. 



192 DIFFERENTIAL AND INTEGRAL CALCULUS. 

PROBLEMS. 

1. Find the area of the logarithmic spiral r=a B , between the 

limits r 2 and i\. 

dr 



Here dr = a log a d6, and dO = 



r log a 



Hence A = V £*» = j^j[%* = [j^J 

2. Find the area described by one revolution of the radius vector 

of the spiral of Archimedes r = aO. A t . f 2n o A2 7/ i 4 rr s a 2 

r Ans. il = i a z B z dQ = 

Vo 3 

3. Find the area of the lemniscate r 2 = a 2 cos 2 6. Ans. a 2 . 

4. Find the area of a loop of the curve r = a cos 2 0. Ans. ^ ira 2 . 

5. Find the entire area of the cardioid r = ail — cos 0). 9 

2 

6. Find the area of a loop of the curve r 2 cos = a 2 sin 3 0. 

Ans. — -2- log 2. 

4 2 s 

Art. 121. Rectification of Plane Curves referred to 
Rectangular Axes. 



By Art. 72, ds 



•+©7* 



in which s represents the length of the arc. 

H— X2 

dx 



-XT> 



eta; (1) 



the limits of integration being the limiting values of #. 

The process of finding the length of an arc of a curve is called the 
rectification of the curve. 

If y be considered the independent variable, the formula is 



■£[ 



\dy 

in which d and c are the limiting values of y. 



dy, (2) 



AREAS AND LENGTHS OF PLANE CURVES. 



193 



If the arc PR, in Fig. 36, is to be rectified, the value of the first 
derivative is found from the equation of the curve, and substituted in 




Formula (1) or (2). If Formula (1) is used, the limits b and a are OB 
and OA respectively ; if Formula (2) is used, the limits d and c are OD 
and OC respectively. 

PROBLEMS. 

1. Rectify the parabola y 2 = 2px. 



Here 
hence 

Therefore 



dy = p. 
dx y' 



S=f(l+ffi<te=jf(p* + y*)Uy 



= ^2±Z +| log (y + Vp* + if) + a (i) 

Here the value of the constant C may be determined by the first 
method of Art. 34. If the arc is estimated from the origin, then S = 
when y = 0, and these values substituted in (1) give 

= |logp+C; 



hence 



C==^\ogp. 



Therefore S = ^2±2 + P w (tL 

2p 2 & V 



+ Vp 2 + y 
V 



194 DIFFERENTIAL AND INTEGRAL CALCULUS. 

which is the length from the vertex to the point which has the ordi- 
nate y. 

Or if the limits of integration are known, for instance, if the length 
from the vertex to an extremity of the latus-rectum is required, then 
the limits are p and 0, and 



S = 



y ^±j/_ 2 +2> i og {y + yy + f) 



ipV2+|log(l + V2). 



2p 2 

2. * Rectify the semi-cubical parabola y 2 = ax 3 . 

27 aV 4 J 27 a 

3. Rectify the curve whose equation is y 2 == — , and determine the 

length of the curve from the origin to the point whose abscissa is 10. 

Ans. 19.0248. 

4. Rectify the circle x 2 -f y 2 = r 2 . 

Here s = 4 f Yl + -\ Klx = 4r f ' dx = 2 wr. 

Jo ^ 2/V Jo Vr 2 - z 2 

But as the result is in circular measure, the circle is a non-rectifiable 
curve. 

An approximate result may be obtained by a series. 



dx , 

— — — — = 4?' 



"a; x? 1 • 3 x 5 

r 2-3r 3 2.4-or 5 



Therefore 



- 4r JT 

2 ri + J^ + -liJ- + 3.5 ; 1 

|_ 2-3 2.4-5 2.4.6.7 J 



From this equation the approximate value of -n- can be determined 
with any required degree of accuracy by taking a sufficient number of 
terms. 

5. Rectify the ellipse y 2 =(1 — e 2 )(a 2 — x 2 ). 



Here *y = -(1 - e 2 f- = - xVl ~ «' • 

dx y Va 2 - x 2 

* The semi-cubical parabola was the first curve whose rectification was effected 
algebraically. (Neil, in 1657, Phil. Trans., 1673.) 



AREAS AND LENGTHS OF PLANE CURVES. 195 



Hence s = 4 C\ H \~ e " x \ lx = 4 f° 

Jo * a 2 — x 2 Jo 



V 



: ( a 2 _ gV)i 



a 2 -* 2 



J° VcT^^V 2a 2.4a 3 "'J 



e 2 3e 4 
== 2 Tea [ 1 — — 



2 2 2 2 • 4 2 



6. Rectify the hypocycloid x 3 -+- y 3 = a 3 . 

Ans. S = § aW ; the entire curve = 6 a. 



7. Rectify the cycloid x = r arc vers - — V2 ry — y 2 . 
Here d * 



fy V2 ?t/ - 2/ 2 
Therefore s = 2 f Y 2r V % = 8 r. 



2/. 

Art. 122. Rectification of Curves in Polar Coordinates. 
By Art. 73 (2), 



■'$ 



f—Q 2 



-[■ 

-IT- 



"ott 



PROBLEMS. 



1. To find the length of the cardioid r = a(l + cos#). 



Here — = — a sin 

eft9 



Therefore s = 2 f "" [a 2 (1 + cos 0) 2 + a 2 sin 2 0]* a*0 

= 2a r r (2+2cos0)^c70 = 4a C" cos f-d$ 

c/0 i/O 2 

0~|rr 

a sin - = 8 a. 
2jo 



196 DIFFERENTIAL AND INTEGRAL CALCULUS. 

2. Rectify the spiral of Archimedes r = ad. 



Ans. ^±f)l 



2a 



+ £lo| 



+ Va 2 + 



3. Rectify the logarithmic spiral logr = # between the limits r x 
and r . Ans. (1 + m 2 y(i\ — ?* ). 



4. Rectify the curve r = a sim 







Ans. 



S-n-a 



Art. 123. The Common Catenary. 

The common catenary is the curve assumed by a flexible cord of 
uniform thickness and density, fastened at two points, hanging freely 
and acted upon only by the force of gravity. 

As the cord is regarded as perfectly flexible, the only force acting 
at any point of the cord is a pull in the direction of the cord at that 
point, which is called tension and is a function of the coordinates of 
that point. 





Y 






\ s 










/? 





N 


/ A 


0' 






A 



Fig. 37. 

In Fig. 37, let 0, the lowest point of the curve, be the origin, and 
let the horizontal line through be the X-axis, and the vertical OY 
be the F-axis. Let (x, y) be any point P on the curve, s the length of 
OP, and c the length of the cord whose weight is equal to the tension 
at O. 

If the weight of the unit of length be taken as the unit of weight, 
the length s will represent the weight of the arc OP, and the length c 
will represent the tension at 0. 



AREAS AND LENGTHS OF PLANE CURVES. 197 

Then the arc OP may be regarded as a rigid body in equilibrium 
under three forces : the tension at P in the direction of the tangent, 
the horizontal tension c at the origin, and the weight s acting verti- 
cally downward. 

Draw PN tangent to the curve, and PS parallel to OX at P. Then 
by the triangle of forces, the sides of the triangle PSN will represent 
the three forces acting on the arc OP. 

rrn n NS weight of OP s 

Therefore = s_ — _ • 

SP tension at c 

hence -^ = — ■ (1) 

dx c v ; 

Differentiating (1), substituting the value of ds, and reducing, 

(2) 



VlxJ dx 



V 



■'* 



Integrating (2) and noticing that when x = 0, -^ = 0, 



H?w i+ (£) 1 



^+\ 1 + 



iwwi 



whence -^ = A I e c — e c ) . 

dx 2 v ' 

Integrating (3), and noticing that x = when y — 0, 
ll = ( kv c + e c ) -<h 



(3) 



(4) 



which is the required equation. 

Removing the origin to the point 0', which is at a distance c below 
0, the equation becomes 



y = |^M e c j- 



198 DIFFERENTIAL AND INTEGRAL CALCULUS. 

In order to rectify the catenary, (5) is differentiated, giving 
f;=iU- e~-), as in (3), 
from which is obtained 

ds =\\e e + <r~ e )d&. 

Therefore s = \ C[e° + e~' c )dx = ^[e c - e~~ c ). 

PROBLEM. 

What is the cnrve in which the cables of a suspension bridge hang ? 

Ans. A parabola. 



CHAPTER XX. 

SURFACES AND VOLUMES OF SOLIDS. 
Art. 124. Surfaces axd Volumes of Solids of Revolution. 

1st. Surfaces. In Fig. 38, let the plane curve MN revolve about 
the X-axis. Let M be a fixed point and P any other point of the curve 
whose coordinates are (x, y). 



M 
/ 


| 


5 Q„ 




T 

y? 




-R 


N 


C 











A B 
Fig. 38. 



Assume MP=s and PQ=As, then the coordinates of Q are 
(x + \x, y + ±y). Let S represent the area of the surface generated 
by the revolution of MP, and A# the surface generated by PQ. Draw 
PR and QT, each equal in length to As, and parallel to OX. In the 
revolution PR generates the convex surface of a cylinder whose area 
is 2-iryks, and QT generates the convex surface of a cylinder whose 
area is 2 tt (y -f A?/) As. Obviously the area of the surface generated bj 
PQ lies between the areas of the cylindrical surfaces. 

Hence 2 uy As < AS < 2 it (y + Ay) As. 

Therefore, as As approaches zero, 

dS = 2 Try ds, 

199 



200 DIFFERENTIAL AND INTEGRAL CALCULUS. 

and S=('2Trij(U (1) 



-°-S'[ 



'+(sy 



dx. (2) 



In like manner for the surface generated by revolving the curve 
about the F-axis, 

The surface of a zone, included between two planes perpendicular 
to the X-axis and corresponding to the abscissas b and a, is 



dy. (3) 



tJj ds: (4) 



2d. Volumes. Let V denote the volume generated by the surface in- 
cluded by the curve MP, the ordinates ML and PA, and the X-axis. 

Let A V represent the volume generated by APQB. 

The volume of the cylinder generated by APGB is m-y 2 Ax, and the 
volume of the cylinder generated by ASQB is ir(y + Ay) 2 Ax. 

•Obviously, iry 2 Ax < A V < it (y + Ay) 2 Ax. 

Therefore, as Ax approaches zero, 

dV= iry 2 dx, 



and V- 



r I y 2 dx, (5) 



In like manner, for the volume generated by revolving the curve 
about the F-axis, 

V=rr Cxhbj. (6) 



PROBLEMS. 

1. Find the surface of the sphere generated by revolving the circle 
x 2 + y 2 = t 2 about a diameter. 

tt / o -9x4 j dy x 

Here y = (r 2 — x- 2 and -±- = 

J v dx y 



SURFACES AND VOLUMES OF SOLIDS. 201 

Therefore S = 2tt Cy(^l + ^ dx = 2Ttr C dx = ±Tri*. 

2. Find the volume generated by revolving the parabola y 2 = 2px 
about the X-axis. 



V = 7T I 2 px dx = p-rrx 2 = i iry 1 • x. 



3. Find the volume of the cone generated by revolving y = x tan a, 
when a is the semi-vertical angle of the cone. 

Ans. V= \ volume of circumscribing cylinder. 

4. Find the volume of the sphere generated by revolving x 2 +y- = ?* 2 
about the X-axis, and also the volume of a spherical segment between 
two parallel planes at distances b and a from the centre. 

Ans. f Trr 5 and tt [r 2 (b - a) - J (6 3 - c 3 )]. 

5. Find the surface and volume of the prolate spheroid generated 
by revolving y 2 = (1 — e 2 )(a 2 — x 2 ) about the X-axis. 

Aiis. S = irb 2 H- - — arc sin e and V= 7rft ■ 
c 3 

6. Find the surface and volume of the right circular cone, gener- 
ated by revolving the line joining the origin with the point (a, b) about 

the XaxlS - Ans. S = „i, V^+l? and V= ««* 

7. Find the surface generated by the cycloid 

y = r arc vers - -f V2 rx — x 2 , 
r 

when it revolves about its axis. Ans. 8 7rr 2 (7r — |). 

8. Find the volume generated by the cycloid 

y 

x = r arc vers - — V2 ry — y 2 , 
when it revolves about its base. Ans. 5 7r 2 r 3 . 

9. Find the surface and volume of the annular torus, generated by 
revolving the circle x' 1 + (y — 5")' 2 = 4", about the X-axis. 

Ans. # = 394.79 sq. in., and V= 394.79 cu. in. 



202 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 125. Surfaces by Double Integration. 

In Fig. 39, let (x, y, z) and (x + dx, y + dy, z + dz) be the coordinates 
of two consecutive points P and E on the given surface whose equation 
is known. Through P and E pass planes parallel to the planes XZ 
and YZ. These planes will intercept an element PE of the curved 
surface, which is projected on the XY-plane in PS = dxdy. 




Fig. 39. 



Let S represent the required area, and dS the area of the element 
PE. 

The area of PS is evidently equal to the area of PE, multiplied by 
the cosine of the angle which PE makes with XY. 

Representing this angle by y, 

area PE • cos y = dx dy ; 
hence area PE = dx dy • sec y. 

By the aid of analytical geometry of three dimensions, 



secy 



f^YJ, 



dx) ^\3yJ 



in which — and — are partial derivatives from the equation of the 
dx dy 

given surface. 



SURFACES AND VOLUMES OF SOLIDS. 203 



Therefore area PE = dS = \l+[ — ) + 



= dS = ["] 



dz\*.fdz 
dx) \dy 



•-/JKSHr 



dxdy, 
dxdy. 



The effect of the ^/-integration, x remaining constant, will be to give 
the sum of all the elements similar to PE from W to M; hence the 
limits of the ^/-integration will be y = CM= VOM 2 —x? and y = 0. 

The effect of the subsequent ^-integration will be to give the sum 
of all the strips similar to WMN forming the given surface ; hence the 
limits of the second integration are x — OA and x = 0. 

Art. 126. Volumes by Triple Integration. 

The given volume is supposed to be divided into elementary 
rectangular parallelopipeds by planes parallel to the three coordinate 
planes ; such an element of volume is represented by kl in Fig. 39. 

The volume of such an elementary parallelopiped is dx dy dz ; hence 
the whole volume is 

V= C f Cdxdydz. (1) 

The limits of integration are obtained from the equation of the 
bounding surface, being so chosen as to embrace the entire volume. 

If the volume included between the three coordinate planes and the 
mrved surface is required, the limits are found as follows : 

The effect of the ^-integration is to sum all the elemental parallelo- 
pipeds in the prism PS; hence the limits of the first integration are 
PE = z =f(x, y) and z = 0. The effect of the ^/-integration is to sum 
all the elemental prisms in the slice WN; hence the limits for the 
second integration are CM=y = f(x) and y = 0. The effect of the 
^-integration is to sum all the elemental slices composing the whole 
volume ; hence the limits of the third integration are x = OA and x = 0. 

X 11 z 

For example, to find the volume of the ellipsoid — + ^- + — = 1, cut 

a- b~ c- 

off by the coordinate planes. 



Here the limits of the ^-integration are c a/1 — • „ and 0, tl 



ie 



204 DIFFERENTIAL AND INTEGRAL CALCULUS. 



Vx 2 
1 and 0, and the limits of the 

^-integration are a and 0. 

VX V I X 
1 — ^-, and yi = b \/l -in the formula, 
cr b z * a 2 

gives for the whole ellipsoid 

V=8 C C H C'dxdydz. 

Jo Jo Jo 

Integrating on the hypothesis that z is the only variable, 

^= 8c XX( 1 -S-S)^ d2/ - 

Integrating again, now on the hypothesis that y is the only variable, 

F= 8c p f y \y*-f)hlxdy = ^ C [y-(y*-tf)i + Vl arc sin ^T'rfs 
b Jo Jo b Jo |_2 2 2/jJ 

Mr*?*'**. 

b Jo 2 2 
Integrating finally with respect to x, gives 

r== 2^ r a ^ 2 _ ^ ^ = ^ ^ a6Ci 

PROBLEMS. 
1 . Find the surface of the sphere x 2 + y 2 -\-z 2 = a 2 . 

Here * = _ * and ^L = _ £ 

dec z oy z 

before 5- //(l + g + f **.//^g*- 



Integrating with respect to ?/, between ?/ = Va 2 — a? 2 and y = 0, 



= /• /v~ attedy = rr ^ arc sin _^ 

J J) Va 2 -a; 2 -y 2 ^ [_ Va 



2/ 



v/o2 12 



SURFACES AND VOLUMES OF SOLIDS. 205 

Integrating with respect to x, between x = a and x = 0, 

b = I -aax = — > 

Jo 2 2 

which is the area of one-eighth of the surface of the sphere. 

2. A sphere x 2 -f- y 2 + z 2 = a 2 is cut by a right circular cylinder 
y 2 z=ax — x 2 . Find the area of the surface of the sphere intercepted 
by the cylinder. Ans. 2a 2 (w — 2). 

3. Find the surface intercepted by two right circular cylinders 
x 2 + z 2 = a 2 and ar -f- ?/ 2 = a 2 . Ans. 8 a 2 . 

4. Find the volume of a right elliptic cylinder whose axis coin- 
cides with the X-axis and whose altitude = 2 a, the equation of the 
base being c 2 y 2 -f b 2 z 2 = b 2 c 2 . Ans. 2 -n-abc. 

5. Find the volume of the solid contained between the paraboloid 
of revolution x 2 -f y 2 = 2 z, the cylinder x 2 -f- y 2 = 4 z, and the plane 2 = 0. 

x2 + j,2 



Ans. 2 C C Vix ~ x2 C ' dxdydz = 37.699+. 

c/0 c/0 i/Q 



6. Find the volume of the solid cut from the cylinder x 2 -{-y 2 = r 2 
by the planes z = and z = x tan a. Ans. -| a 3 tan a. 

7. Find the entire volume bounded by the surface x* 4- y* + 33 
= ^25. .4ns. 44.88. 



CHAPTER XXI. 

CENTRE OF MASS. MOMENT OF INERTIA. PROPERTIES OF GULDIN. 
Art. 127. Definitions. 

The definitions of this article are taken from Mechanics and are 
here assumed without investigation. 

The moment of any force with respect to an axis perpendicular to 
its line of direction is the product of the magnitude of the force by 
the perpendicular distance from its line of direction to the axis. The 
moment of a force with respect to a plane parallel to its line of direc- 
tion is the product of the force by the perpendicular distance from its 
line of direction to the plane. 

The force exerted by gravity on any body is proportional to the 
mass of the body, and hence the mass of the body may be taken as the 
measure of the force exerted on it by gravity. 

The centre of mass of a body is that point so situated that the 
force of gravity produces no tendency in the body to rotate about any 
line passing through the point ; hence it may be regarded as the point 
at which the whole weight of the body acts. The centre of mass is 
sometimes called centre of gravity and centre of inertia. 

The moment of inertia of a body with reference to a straight line, 
or plane, is the sum of the products obtained by multiplying the mass 
of each element of the body by the square of its distance from the 
line or plane. 

Points, lines and surfaces, as here considered, are supposed to be 
material bodies. Lines, surfaces and solids are regarded as being com- 
posed of an infinitely large number of indefinitely small particles. The 
weight of a body is the resultant of the weights of all of its elemental 
particles acting in vertical lines, and the resultant of this system of 
parallel forces passes through the centre of mass. 

206 



CENTRE OF MASS. 



207 



Art. 128. General Formulas for Centre of Mass. 

Assume a system of rectangular coordinate axes, retaining a fixed 
position with reference to the body, the plane XY being horizontal. 
Let a small particle of mass at any point (x, y, z) be represented by 
Am. Then the force exerted by gravity on Am is measured by Am in 
a direction parallel to the Z-axis. 




Fig. 40. 



If the mass of Am were concentrated at the point (x, y, z), the 
moment of the force exerted on Am with respect to the plane YZ 
would be xkm ; and the sum of the moments of all the elements of the 
body with reference to this plane would be 2#Am. 

The resultant force of gravity is 2A??i, and if the coordinates of the 
centre of mass be represented by (x, y, z), as the centre of mass is the 
point through which the resultant passes, x^\m will be the moment of 
the resultant with reference to the plane YZ. But by the principle of 
moments, the moment of the resultant of any number of forces is equal 
to the algebraic sum of the moments of the forces. 

Hence, x%\m = 2£.rAm. 

If now Am diminishes indefinitely, 



Therefore 



x I dm = I x dm. 

I xdm 
x =- . 

fdm 



(1) 



208 DIFFERENTIAL AND INTEGRAL CALCULUS. 

I ydm 

Similarly, y =^— , (2) 

I dm 

Jzdm 
(3) 

J dm 

The mass of any homogeneous body is the product of its volume 
by its density. If ~k represents the constant density and dv the ele- 
ment of volume, then kdv = dm, and (1), (2) and (3) become 

I xdv 

*=V~,- (4) 

Jdv 

( y dv 

y=' L r -> ( 5 ) 

jdv 

I zdv 

and z= ^— — , (6) 

jdv 

If the body is a material line in the form of the arc of any curve, 
and if ds is the length of an element of the curve, Formulas (4), (5) 
and (6) become 

/xds 



(7) 



ds 



f- 

[yds 

y=^-^> (8) 

fds 

I zds 
z=J—. (9) 

fas 

If the curve is a plane curve, it may be taken in the plane XY, in 
which case z will be zero. 



CENTRE OF MASS. 



209 



PROBLEMS. 



1. Find the centre of mass of an arc of a circle, taking the diam- 
eter bisecting the arc as the X-axis and the left vertex as the origin. 
In Fig. 41, let AOB be the arc. 




Fig. 41. 



The equation of the circle is y 2 = 2 ax — x 2 ; 



hence 



dy 



(a — x) dx 
V2 ax — x 2 



ds = Vdx 2 + dy 2 = 



adx 



Therefore x 



fds sJ ° 



V2 ax — x 
xdx a 



V2 ax — x 2 s 



( — V2 ax — x 2 + s) = a — 



ay 



2. Find the centre of mass of an arc of the hypocycloid 



x s _|_^3 —a*, 



between two successive cusps 
Here 



hence 



• ( - ) * dx 



ds = Vdx 2 + dy 2 = y* v x I _|_ y I dx 



-\/ 2 2 / - 2 1 

= v « 3 — xtyjx 3 + - 1 — 



dx 



a* — x-- 



dx. 



210 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Therefore 



OM=x 



i" X {tt dX 



f©* 



= f 



Similarly, MG = y=*a. 




Fig. 42. 



3. Find the centre of mass of the arc of a semi-cycloid. 

Ans. x = (tt — f ) a, y = 

Art. 129. Centre of Mass of Plane Surfaces.* 
If rectangular coordinates are used, 
dv = clA = dxdy, 
and Formulas (4) and (5) of Art. 128 become 

I I xdxdy 
x = '' ? ; -, 

jjydxdy 



and 



I I tfccct?/ 



a) 



(2) 



* The centre of mass of a plane area is sometimes called the centroid of the 



CENTRE OF MASS. 



211 



PROBLEMS. 



1. Find the centre of mass of the area included between the parab- 
ola y 2 = 2 px and the double ordinate whose abscissa is a. 



r*a s*y r*a 

I I xdxdy I xydx 

. Jo J-y _ c/o 

dxdy I ydx 
y Jo 

■\/2px^dx 



2px z dx 



2. Find the centre of mass of the semicircle x 2 -\-y 2 = r 2 on the 
right of the F-axis. 



I I xdxdy I xydx 

-__ J0 J-y _J0 

I J dxdy I ydx 

I x^/r 2 — 



x 2 dx 



x 2 dx 



4r 
3x 



y=o. 



3. Find the centre of mass of an elliptic quadrant whose equation 



is y 



a 



x\ 



A - 4a - 4 5 

Ans. x = — , y = 

3tt J 3tt 



4. In Fig. 43, ABD is a segment of a parabola cut off by an ordi- 
nate, and BE is parallel to Ax. 

1st. Determine the distance of the centre 



of mass of ABD from Ax. 



Ans. 



3y 



fd. Determine the centre of mass of 



ABE. 



Sx 3?/ 



10 4 

5. Find the centre of mass of the cycloid. 




Ans. x — wr, y = 4- 



212 DIFFERENTIAL AND INTEGRAL CALCULUS. 

6. Find the centre of mass between y m = x n and y n = x m . 

Ans. x = y = 5b — Z, — I 

(m + 2fi)(2m + »i) 



Art. 130. Centre of Mass of Surfaces of Revolution. 

If a curve in a plane with the X-axis be revolved about this axis, then 

dv = 2 7ryds', 
hence, by Art. 128 (4), 

J 2 irxy ds I xy ds 

x=^ =^~ 

J 2 iry ds I yds 

PROBLEMS. 

1. Find the centre of mass of the convex surface of a right cone, 
generated by the line y = ax. 



Here ds = ^/dx 1 -f dy 2 = Vcr -\-ldx 

I ax 2 Va 2 -\-ldx 
I ax V<x 2 -+- 1 dx 



hence x = ^ — — \ x 



2. Find the centre of mass of the surface generated by the revolu- 
tion of a semi-cycloid x = a vers -1 - — V2 ay — y 2 about its base. 



Here ds = 



a 

V2 a dy 
V2a — y 



f 2a yydy 

hence x = — ^- - — - = ft a. 

C 2a y dy x ° 

J° V2 a — ?/ 

3. Find the centre of mass of the convex surface of a hemisphere 
whose radius is equal to 10. Ans. x = 5. 



CENTRE OF MASS. 213 



Art. 131. Centre of Mass of Solids of Revolution. 

If a solid be generated by the revolution of a plane curve about 

the X-axis, then 

dv = 2irydydx\ 



hence, by Art. 128 (4) 



Ci xydxdy 
J J ydxdy 



PROBLEMS. 

1. Find the centre of mass of a right circular cone, whose convex 
surface is generated by revolving y = ax about the X-axis. 



nax 

ydxdy 

Jo Jo 

f 



arx* 7 
dx 

2 -*. 



Jo 2 

2. Find the centre of mass of a paraboloid generated by y 1 — 4 ax. 

n ax xydxdy 


I P " z ydxdy 

Jo Jo 

J 2 ax?dx 
_0 =2-X 

2 a# c?# 

3. Find the centre of mass of a hemispheroid generated by 

y 2 = — (2 a<c — a; 2 ). ^4ws. | a. 



214 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 132. Moments of Inertia of Surfaces. 

In Fig. 44, the curves AB and CD and the ordinates LN and PM 
intercept a plane surface PLSR, whose moment of inertia is required. 

The surface is supposed to be divided into rectangular elements by 
lines parallel to the coordinate axes. 



f 


D 






c 


) - 


r 








L 


A""^ 




















^"**^ 




















""^B 


























J 


































C^^ 


R , W V 






^^-D 



Fig. 44. 



Let (x, y) be the coordinates of any point as /, then (x + dx, y -f dy) 
will be the coordinates of g, and dxdy will be the area of the element fg. 

The moment of fg about X— y 2 dxdy. 

Let the equations of AB and CD be y = f(x) and y = 4> ( x ) respec- 
tively ; and let ON = b and M = a. 

If x be regarded as constant, while y varies from <j>(x) tof(x), the 
integration will give the moment of the vertical strip WQTV. 

Then in the second integration, x varying from a to b, the sum of 
the moments of all the strips composing the area PRSL will be given. 

Representing the moment of inertia by M. I., 

y 2 dx dy. 



M.I. 



<*>(*) 



PROBLEMS. 

1. Find the moment of inertia of a circle about its diameter. 



M 



I. = J | yhlxdy 

«y -r\) —\/r n -~ xl 



PROPERTIES OF GULDIN. 215 

2. Find the moment of inertia of a rectangle about an axis through 
its centre parallel to one of its sides. 

Let 2 b and 2 d denote the width and length respectively, the axis 
being parallel to b ; then 



M. I. = C f d y 2 dx dy = ± bd\ 
J-hJ -a 



3. Find the moment of inertia of an isosceles triangle about an 
axis which passes through its vertex and bisects its base. 

Let a = the altitude and 2b = base, and take the origin at the 
vertex and the axis of moments as the X-axis ; then 



-X 

M. I. = f ° C\ dxdy = \ db\ 



Art. 133. Guldin's Theorems.* 

I. Let a plane curve in the same plane with the X-axis revolve 
about the X-axis. 

The ordinate of the centre of mass is 



fas' 



by Art. 128 (8). 



Therefore 2 Try • s = 2 it Cy ds. (1) 

But by Art. 124 (1), the second member of (1) is the area of the 
surface generated by the revolution of the curve whose length is s 
about the X-axis, and the first member is the circumference described 
by the centre of mass, multiplied by the length of the curve s. 

Hence, if a plane curve revolve about an axis in its own plane 
external to itself, the area of the surface generated is equal to the 
length of the revolving curve, multiplied by the circumference de- 
scribed by its centre of mass. 

* Sometimes called Theorems of Pappus, as they were first stated by Pappus. 



216 DIFFERENTIAL AND INTEGRAL CALCULUS. 

II. A plane area revolves about the X-axis. The ordinate of the 
centre of mass of the plane surface is 

I I ydxdy 

V-~i-f. ' by Art. 129 (2). 

I J dxdy 

Therefore 2-irjj y i dxdy = 2?r I (ydxdy == tt J ifdx. (2) 

But by Art. 124 (5), the last member of (2) is the volume gener- 
ated by the revolution of the area ; and in the first member, I I dxdy 
is the revolving area. Hence, if a plane area revolve about an axis 
external to itself, the volume generated is equal to the area of the 
revolving figure, multiplied by the circumference described by its 
centre of mass. 

If the curve or area revolve through any angle 6 instead of making 
an entire revolution, 6 must be substituted for 2tt in equations (1) 
and (2). 

PROBLEMS. 

1. Find the surface and volume of the ring generated by revolving 
a circle whose radius = r, about an external axis distant b from the 
centre of the circle. Ans. S = 4:ir 2 ab, V— 2ir 2 a 2 b. 

2. Find the volume generated by an ellipse revolved about an axis 
distant 10 from the centre ; the semi-axes being 10 and 5. 

Ans. 9869.6+. 

3. Find the surface and volume generated by revolving a cycloid 
about its base. Ans. <S = -^ -rra 2 , V — 5 7r 2 a 3 . 



CHAPTER XXII. 

DIFFERENTIAL EQUATIONS. 

Art. 134. Definitions. 

A differential equation between two variables x and y is an equation 
containing one or both of the variables x and y and one or more deriva- 
tives, such as -^, — \ — ^, etc. 
dx dxr dx 6 

The order of a differential equation is that of the highest derivative 
which it contains. 

The degree of a differential equation is that of the highest power 
to which the highest derivative which it contains is raised. 

The solution of a differential equation consists in finding a relation 
between x and y and constants, from which the given equation may be 
derived by differentiation; this relation is called the primitive. The 
solution requires one or more integrations, and each integration intro- 
duces an arbitrary constant ; hence the solution of a differential equa- 
tion of the nth order will give an equation containing n arbitrary 
constants. 

The same primitive may have several differential equations of the 
same order. 

For example, given the equation 

ay + bx + c = 0. (1) 

(2) 
(3) 

ax 

217 



By differentiating, 


( 


dx 


f& = 


= 0. 


Eliminating a between 


(1) and 


■(2), 








dx 


dy 

dx 


by-- 


= 0. 


Eliminating b between (1) and 


(2), 








ay-\-c 


— ax 


dx 


= 0. 



218 DIFFERENTIAL AND INTEGRAL CALCULUS. 

In this example, equation (1) is called the complete primitive, and 
equations (2), (3) and (4) are differential equations showing the same 
relation between the variables. 



Art. 135. Differential Equations of the First Order and 

Degree. 

The general form of the differential equation of the first order and 

degree is 

Mdx + Ndy = 0, (1) 

in which M and N are functions of x and y. This equation may be 
put in the form 

dx 

The most obvious method of solving a differential equation of the 
first order and degree is by means of the separation of the variables, 
whenever practicable. The variables are separated when the coefficient 
of dx contains the variable x only, and the coefficient of dy contains the 
variable y only; that is, when the equation can be reduced to the form 

Xdx + Ydy = 0, 

in which X is a function of x only, and Y is a function of y only. 
Let the form be 

XYdx + X'Y'dy = 0, (2) 

in which X and X' are functions of x only, and Y and Y' are func- 
tions of y only. 

Dividing by X'Y, 

^ + 1^ = 0, (3) 

in which equation the variables are separated. 
For example, given 

(1 - xfy dx - (1 + y) x 2 dy = 0. 
Dividing by x*y, 

x 2 y 



DIFFERENTIAL EQUATIONS. 219 

Hence, «*-?& + & -$L-dy = 0. 

x 2 x y 

Integrating, 2 log x + x — log y — y = G. 

x 

PROBLEMS. 

1. (1 — y) dx + (1 -f- x) dy = 0. Ans. log (1 + x) - log (1 — y) = G. 

2. (1 + y 2 ) dx — x^dy = 0. Ans. 2 x% — arc tan y = G. 

3. (1 + x)ydx-\- (1 — y)xdy = 0. ^4?is. log (r?/) + £ — y = 0. 

4. dy -\-y tan a; da; = 0. Ans. log ?/ — log cos x = (7. 

. dy 1 + y 2 a x + G 

7. sin x cos y dx — cos x sin ydy = 0. Ans. cos ?/ = (7 cos ic. 

8. Hemholtz's equation for the strength of an electric current C at 

rp T flG 

the time t is (7 = — , in which E, R and L are given constants. 

R R dt 

Find the value of C, determining the constant of integration by the 
condition that its initial value shall be zero. 

9. The equation showing the strength of current i for the time t 

dC 
after source of E. M. F. is removed, is RC = — L — (R and L being 

constants). Find the value of G 

Ans. C = Ie L , in which I = current when t = 0. 
10. The differential equation of the current of discharge from a 

condenser of capacity G in a circuit of resistance R is — = — — • 

i GR 

t 

Find i, if the initial current is Jo- 



Art. 136. Homogeneous Differential Equations. 

The differential equation Mdx + Ndy = is said to be homo- 
geneous when M and N are homogeneous functions of x and y of the 
same degree. 



220 DIFFERENTIAL AND INTEGRAL CALCULUS. 

If the equation is written in the form 

dy_ _M 

dx~ N' 

y 

the second member is seen to be a function of — 

x 

y 

If, now, v be substituted for -, 

x 

-, dy dv , 

y = vx, and -^ = x — • -f- v, 
dx dx 

and the equation becomes ■ 

in which the variables can be separated, giving 

dx _ dv 

x f(v) - V 

For example, given the homogeneous equation, 

2 . ody dy 

y l -f- ar-^ = xy -^- 

dx dx 

Substituting y = vx, 

2 2 x 2 (x dv + v dx) _ vx 2 (x dv -f v dx) m 
dx dx 

, dv . dx , 
whence 1 = dv. 

v x 

y 

Integrating and substituting v = '-, 

x 

y y 

log- + log a; = - + c. 

y 
Therefore log y — c = -, 

or c 1 e i =2/.(logc 1 = c). 



PROBLEMS. 

x 

1 . (x — 2y)dx + ydy = 0. Ans. log (y — x) = c. 

y — x 

2. (2Vxy — x)dy + y dx = 0. Ans. y — ce~\y. 



DIFFERENTIAL EQUATIONS. 221 



3. x 2 dy — y 2 dx — xydx = 0. Ans. log x + - = c. 

y 



y 



4. xdy — ydx= dx^Jx 2 — y 2 . Ans. log x = arc sin - + c. 

5. (8 y + 10 a;)da; + (5 # + 7a:)efy = 0. Ans - (y + x) 2 (y -{-2x) s = c. 

6 . (.t 2 + y 2 ) dx — 2xydx = 0. Ans. x 2 — y 2 = ex. 

7. Find the curve in which the subtangent is equal to the sum of 
the abscissa and ordinate at any point of the curve. 

From Art. 70 (3), it is seen that the differential equation is 

dx 
-y — =x + y; 

dy 

whence y 2 -f- 2 xy = c, a hyperbola. 

Art. 137. The Form (ax + by + c)dx + (a'x + &'y -f c')dy = 0. 

The equation Mdx -f- Ndy = can always be solved when M and N 
are functions of x and ?/ of the first degree, or having the form 

(ax + by + c)dx + (a'x + b'y + c') dy = 0. (1) 

Assuming x = #' + ft and y = y' + k, and substituting in (1), 

(ax' + by' + a/i + 6A: -f c) cfa f + (a'x' + &'#' + a'ft + b'k + c')dy' = 0. (2) 

In order that (2) may be homogeneous, 

ah + bk + c = 0, and a'/z. -f 6'ft + c' = 0, 

-, cb' — c'b A -, ac' — a'c 

giving h = — -, and k = — -• 

a'o — ab' a'b — ab' 

Equation (2) now becomes 

(ax' + by') dx' + (a'a' + b'y')dy' = 0, 

a homogeneous equation, and the variables can be separated as in the 

preceding article. 

a' b' 
This method evidently fails when a'b = ab' ; that is, when — = — 

a' b' a b 

In this case put — = — = m, a' = ma, b' = mb. 
a b 

Equation (1) now takes the form 

(ax + by + c) dx -\- [m (ax -f- by) + c'] aj/ = 0. (3) 



222 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Assuming ax + by = z, dy = — ^-^ — , 

and substituting in (3), 

(z + c)dx + (mz + c') — — = ; 

whence [b (z + c) — a (mz — c')] cto + (mz + c') cfc: = 0, 

and dx + ("■« + «')<»* = , 

& (z + c) -— a (mz — c') 

and the variables are separated. 



PROBLEMS. 

1 . (1 + a + y) dx + (1 4 2 a; + 3 y) dy = 0. 
Assuming x = x' + ft, and y-—y'-{-k, 

(l + x' + h + y' + k)dx' + (l+2x' + 2h + 3y' + 3k)dy l = 0, 
in which ft + ft + 1 = 0, and 2h + 3k + l = 0, 

giving h = — 2, and 7c — 1. 

The equation now becomes 

(0' -f y) (te' + (2a*' + 3y>dy'=.0, 
which is homogeneous and can be solved by Art. 136. 

2. dy = ax+by + c. Ans. abx + b 2 y + a + bc= Ce bx . 
dx 

3. (2x + y + l)dx + (kx + 2y — l)dy = 0. 

Ans. x + 2 y + log (2 x + y — 1) = C. 

4. (2«-?/ + l)(^4-(22/-^-l)% = 0. 

-4ws. a; 2 — xy + y 2 -f x — y = C. 

Art. 138. The Linear Equation of the First Order. 
The equation of the form 

f x + Py=Q, (1) 

in which P and Q are functions of x only, is called a linear equation 
because it is of the first degree with respect to y and its derivative. 



DIFFERENTIAL EQUATIONS. 223 

This linear equation admits of a general solution. As the second 
member is a function of x only, an integrating factor of the first 
member will be an integrating factor of the equation if it is a function 
of x only. To find such a factor, put y = Xz, in which X is an arbi- 
trary function of x, and 2 is a new variable. Then dy = Xdz + zdX, 
which reduces (1) to 

Xdz + z dX + PXz dx = Qdx. (2) 

Assume zdX= Qdx, (3) 

then (2) becomes Xdz + PXz dx = ; 

whence — = — Pdx ; hence log 2 = — J Pdx, and 2 = e~^ dI . 

Substituting this value of z in (3), 

e^ pdx dX = Qdx, or dX= e$ Pdx Qdx. 

Therefore X = fe$ Pdx Qdx + c, or y = Xz = e~f Pdx ( CeS Pdx Qdx + c\ 

PROBLEMS. 

' r, yxdx a , 

1 + x- 1 + ar 

This linear equation for ?/, put in the general form, is 

dy yx a 

dx ~ 1 + x 2 ~ 1+a/ 



in which I Pdx = — | — - — = log 

J J 1 + x 2 5 A /Tn 



+ ar Vl+x 2 

Hence the integrating factor is 

e fp*z = e toga+* 1 )-* = (1 + x 2 )~i 

p-Qcto = f^a±^!)z dx = ™ + c . 

J J 1 + ar (1-fz 2 )* 

Therefore y = (1 + a? 1 )* ( — — + (A= cue + c(l + aM 

V(i + ^ ; 



224 DIFFERENTIAL AND INTEGRAL CALCULUS. 

2. x 2 ^+(l-2x)y + x 2 . Ans. y = x 2 (l+ce*\ 

dx 

3. a?— — ay = x 4- 1. -4fts. y = cx a 

dx a — 1 a 

4. Va 2 -f- x 2 — Va 2 + x 2 -+ = x-\-y. 

dx 

Ans. y = a2] °g( x + V° 8 + ^ 2 ) + c - 
a + Vet 2 -f x 2 

5. — cos x + 7/ sin x = 1. Ans. y = sin x + c cos x. 
dx 

6. The differential equation of electromotive forces in a circuit of 
resistance R and induction L, when the impressed E. M. F. is a sinu- 
soid given by the equation e = E sin pt, is i£ sin pi = Ri -\- L— . Solve 
the equation for i, i and t being the only variables. 

Ans. i = ce L -\ - sin (pt — d>), in which <h = tan -1 -^» 

7. The equation of energy for a circuit containing resistance and 
capacity is eidt = i 2 Rdt + (idt) I &c&. Solve the equation for i, when 
e = E sinpZ. 

,4ns. i = — ; — -=- sin ( pt -f- tan -1 — - ) -f ce * c * 

VMS' v **> 



Art. 139. Extension of the Linear Equation. 

The more general form, JL-\.py = Qy" n\ 

dx 

in which P and Q are functions of x only, is readily put in the form of 
the linear equation of the preceding article. 

Dividing (1) by y», I f&+ JL = Q (2) 

y n dx y n ~ l 

Assume 

n 

z = y~ n+l ; whence y n = z n ~ 1 , y n ~ l = z~ l , and dz = — (n — l)y~ n dy. 



DIFFERENTIAL EQUATIONS. 225 

Substituting in (2) and reducing, 

*_(« _!)*), = - (»-l)ft 

ax 

which is the form of the linear equation of the preceding article. 







PROBLEMS. 


1. dy + ydx = xfdi 








Dividing by y^dx, 


1 dy 

y s dx 


& 


-X. 


Assume 




z = 


= 2T 2 - 


Substituting and reducing 


) 






dz _ 


-2z = 


= -2x. 



dx 
By the preceding article, 

Cpdx = -2x, and 

z = _ 2 e 2x Ce- 2 * x dx = e 2 * (xe~ 2 * + ± e - 2x + C). 
Therefore 1 = y 2 (Ce 2x + \ + x). 

2. <M. = x hf - xy. Ans. - ■ = x 2 + 1 + Oe x \ 
dx y- 

3. 3y 2 ^-- ay 3 = x -f 1. Ans. f = Ce ax - £±1 _ A. 

dx a a- 



</ 2 



4. -^ (#y + xy) = 1. ylns. <c 



(2-?/ 2 ) e 2 + (7 



Art. 140. Exact Differential Equations. 

The equation Mdx + JVcfa/ = (1) 

is an exact differential equation, when 

dM = dN (2) 

dy dx' K J 



226 DIFFERENTIAL AND INTEGRAL CALCULUS. 

When the condition in (2) is fulfilled, the integral may be obtained 
by finding J Mdx, regarding y as constant and adding an arbitrary func- 
tion of y. 

The undetermined function of y may be found by the condition that 

the differential of the result just obtained regarding x as constant must 

equal Ndy ; that is, 

±fMdx + V^l = N, 
dyJ dy 

from which f(y) may be obtained. (See Art. 49.) 



PROBLEMS. 



1. ^ + (2y-* 2 )t 

y V y 2 



— = -- and ^"=-1- 

dy y 2 ' dx y 1 ' 



condition (2) is fulfilled, and the equation is exact. I Mdx, treating y 



as constant and adding f(y), is - +f(y). 



d (x\ | df(y) = 2 x 

dy\yj dy y 2 ' 



Now 
whence 

and 

hence 

Therefore 

y 

2 . (6xy- y 2 ) dx + (3x 2 -2 xy) dy = 0. Ans. 3 x 2 y - y 2 x = 0. 

3. x (x + 2 y) dx + (r 2 — ?/ 2 ) dy = 0. ^Ins. ar 5 + 3 x 2 y — y 3 = 0. 

4. (x 2 -4,xy-2 y 2 ) dx + (y 2 -±xy-2 x 2 ) dy = 0. 

Ans. x 3 — 6 cc 2 ?/ — 6 a??/ 2 + 2/ 3 = 0- 

5. aria; + ydy + xdy ~ ydx = 0. Ans. x 2 + ?/ 2 - 2 arc tan 2 = C. 

x 2 -\-y 2 x 

6 . e x ^2 + y 2 + 2 g.) ^oj + 2 ye'cft/ = °- ^ s - e * <>* + 2/ 2 ) = <?• 





+ df(y). 

dy 


= 22/- 


X 




df(y). 
dy 


= 22/; 






f(y)- 


=2/ 2 . 






-+f- 


= 0. 





DIFFERENTIAL EQUATIONS. 227 

Art. 141. Factors Necessary to make Differential Equations 

Exact. 

When Mdx + Ndy is not an exact differential, it may often be trans- 
formed into an exact differential by the introduction of a factor con- 
taining x or y or both. This factor, which converts a given differential 
equation into an exact differential equation, is called an integrating 
factor. 



I. When Mdx + Ndy is homogeneous. 
Mdx + Ndy = \ 



(Mx + Ny)(— + ^\ + (Mx - Ny){ - 



dx _ dyX 
\x y J u '\x y) 



\[{Mx 



+ Ny) d log (xy) + (Mx - Ny) d log - 

yj 



(i) 



(2) 



Hence ^±my = x ( Ife-jp, ^ * 

Mx + Ny 2 v y 2 ilfa; + iVfy b y 

= i d[log,4-logy 2 ] + i^^^when^^ 
Jf £ + Ny v y 

When 3f and N are homogeneous, 

Mx + Ny " w 

and the second member of (2) is an exact differential. 

Therefore is an integrating factor. 

Mx + Ny 

This method fails when Mx + Ny = 0, but in this case Mx = — Ny. 

Dividing the first term of Mdx + Ndy = by Mx, and the second term 

by its equal, — Ny, 

dx_cly = () ^ 

x y 

Therefore y = Cx. 

For example, given (xy + y 2 ) dx — (# 2 — xy) dy = 0. 

1 1 



Here 



Mx + Ny 2 xy 2 



228 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Multiplying by this factor, 

2\y xj 2\y 2 yj 

The condition of integrability for an exact differential is now ful- 
filled. 

Therefore - -f- log (xy) = G. 

II. The form, f 1 (xy)ydx+f 2 (xy)xdy = 0. 

By a method similar to that of I., it may be shown that 



Mx — Ny 

is an integrating factor. This fails when Mx — Ny = 0, but it can then 
be shown, as in the corresponding case of L, that the solution now is 
xy = C. 

But another method of solution by the separation of the variables 
may be used. For example, given 

(x 2 y 2 + xy) y dx 4- (x 2 y 2 — 1) x dy = 0. (3) 

Assume xy = v, 

then (v* + v)-dx+(v*-l)x(--~\=0 } 

X \ X X J 

whence — = — f v ~ ) dv, (4.) 

X \ V J 

and the variables are separated. 

Integrating (4), and substituting v = xy, 

y = ce xy . 

III. To determine the factor necessary to render Mdx + Ndy exact, 
when that factor is a function of one variable only. 

Assume X, a function of x only, to be the required factor; then 
XMdx + XNdy is an exact differential. 

Hence — (XM) = A (XN) ; 

dy ox 

and since — — = 0, 

ay 



DIFFERENTIAL EQUATIONS. 229 

x dM =x dN +N dX. 

dy dx dx 

therefore «a^-MW (5) 

X N\dy dx) W 

The first member of (5) does not contain y; hence the second 
member must be independent of y also, or 

JSr\dy dx J JKJ 

Integrating (5), logX= (f(x)dx, 

therefore X = e$ Ax)dx . 

Similarly, if Y, a function of y only, is the original factor, 

YMdx+ YNdy 
is an exact differential, and 

For example, given (x 2 + y 2 + 2 x) dx + 2ydy — 0. 
Assume X to ,be the integrating factor. 

dy dx 

N\dy dx J 
log X = J dx = x, and X = e x . 
Multiplying the given equation by e x , 

e x (x 2 + y 2 + 2 x) dx + 2 e x y dy = 0. 

The condition of integrability for an exact differential is now ful- 
filled, and integrating as in Art. 140, 

e x x 2 -f- e x y 2 = c. 



230 DIFFERENTIAL AND INTEGRAL CALCULUS. 

PROBLEMS. 

1. x(x 2 + 3y 2 )dx + y(y 2 + 3x 2 )dy = 0. Ans. x 4 + 6 x 2 y 2 + y* = c. 

7 , j , xcly — y dx A x 2 4- y 2 , .V 

2. ydy + xdx-\ ^ — ^- — ^bis. — -!-^- + arc tan ^ = c. 

x 2 + y 2 2 x 

3. (x 2 + ?/ 2 ) dx — 2xydy = 0. ^.ws. x 2 — y 2 = ex. 

4. (x 2 -f 2 #?/ — y 2 ) dx = x 2 — 2xy — y 2 dy. Ans. x 2 + y 2 = c(x + y). 

5. (x + 2/) 2 — = a 2 . Ans. y — a arc tan :1jL^ = c . 

dx a 

6. a^cto + (3 x 2 y + 2 y 3 ) dy = 0. ^bis. # 2 + 2 ?/ 2 = c s/x 2 + 1/ 2 . 

7 . (1 + xy) y dx + (1 — ay) xdy = 0. Ans. x = cye xy . 
8 - (y + 2/ V##) efcc + (# + x^/xy) dy = 0. -4ws. cm/ = c. 

9. (3 # 2 — i/ 2 ) — = 2 cm/. ^4ns. x 2 — y 2 = cy 3 . 

(XX 

10. 2xydy = (x 2 + y 2 ) cfo. .4ns. x 2 — y 2 = ex. 

Art. 142. First Order and Degree with Three Variables. 

The general form of the differential equation of the first order and 
degree between three variables is 

Pdx + Qdy + Rdz = 0, (1) 

in which P, Q and R are functions of x, y and z. 

Such an equation sometimes admits of solution by the separation of 
the variables. 

To obtain the condition of integrability, represent the function by 

v=f(x,y,z) = c>, (2) 

whence — dx-\ dy -\ dz = 0. (3) 

dx dy dz 

Comparing (1) and (3), it is seen that 

P, Q and R are proportional to — , — and — 

dx dy dz 



DIFFERENTIAL EQUATIONS. 231 

To obtain the required relation between P, Q and M, assume the 
factor u such that 

and uB = — - (6) 



From (4) and (5), 



9?/ 3?/ 9a; 9x* 



hence t* — + — — = it -^ + Q — , 

oi/ ay aa; die 

SWlarl* .gfl-*"}-**'-** (8 ) 

-, /3i2 dP\ r>du -edit, /nN 

and "U?-*h P ^ aT (9) 

Multiplying equations (7), (8) and (9) by B, P and Q, respectively, 
and adding, 

Equation (10) is the required condition of integrability. When 
this condition is fulfilled, equation (1) may be integrated by regarding 
one of the variables, x, y or z, as constant, and omitting the correspond- 
ing term, Pdx, Qdy or Pdz. 

Thus omitting Edz, integrating Pdx -f Qdy = 0, regarding z as con- 
stant and introducing / (z) as the constant of integration, .the integral 
is obtained so far as it depends upon x and y. Finally, by comparing 
the total differential of this result with equation (1), df(z) is found in 
terms of z and dz, and then by integration the value of f(z). 

When certain terms of the equation form an exact differential, the 
remaining terms must also be exact. It follows that if one of the vari- 



232 DIFFERENTIAL AND INTEGRAL CALCULUS. 

ables, say z, can be completely separated from the other two, so that in 
equation (1) R becomes a function of z only, and P and Q functions of 
x and y only, the terms Pdx + Qdy must be thus rendered exact if the 
equation is integrable. 

For example, given zy dx — zx dy — y 2 dz = 0. Dividing by y% which 
separates z from x and y, and puts it in the exact form, 

y dx — x dy _ dz _ ^ 
y 2 z 

of which the integral is x = y log cz. 

PROBLEMS. 

1. (x — 3 y — z) dx + (2 y — 3 x) dy + {z — x) dz = 0. 

Ans. x 2 -\-2y 2 — 6xy — 2xz + z 2 = C. 

2. yz 2 dx — z 2 dy — e z dz = 0. Ans. yz = e x (1 + cz). 

3. yzdx-\-zxdy + ?/# dz = 0. ^.ns. ;w/z = C 

4. (?/ efcc + a; dy)(a -\-z) = xy dz. Ans. xy = c (a -f z). 

5. The increase in energy of a magnetic field caused by the inde- 
pendent increments di x and di 2 of current in two coils of self-induction 
L x and L 2 and mutual induction fx is dW= L 1 i 1 di 1 + L 2 i 2 di 2 + fihdi 2 
+ fi^di^ Find the energy of the field. 

Ans. W= J[ ^-h^-hfii 1 i 2 . 



Art. 143. First Order and Second Degree. 

The general form of a differential equation of the first order and 
second degree is 

C -f 2 + M c ^ + N=0, (1) 

dx 2 dx 

in which M and N are functions of x and y. The direct differential 
obtained from the primitive contains only the first power of i -^-, and 

hence cannot be identical with (1). But if the primitive is supposed 
to contain the first and second powers of a constant c, and is solved 
with reference to c, there will result two values of c, from each of which 
c will disappear on differentiation ; and each of these resulting differ- 



DIFFERENTIAL EQUATIONS. 233 

ential equations will contain only the first power of — , each being a 
factor of (1). Hence the product of the two equations will give (1). 

Therefore the given equation is solved as a quadratic in -^, all the 

dx 

terms are transposed to the first member, and it is resolved into two 
factors of the first order and degree. Each of these factors is then 
placed separately equal to zero and integrated, using the same arbitrary 
constant in each. When all the terms in each of these results are 
transposed to the first member, the product of these first members 
placed equal to zero will be the complete primitive. 

For example, given y %L + 2 x ^ - y = 0. (1) 

1 ' 8 u dx 2 dx u w 



Solving for -^-> 
dx 




and 


dy _ 
dx 




dy = _ 

dx 


x yV + y 2 
— r > 

y y 


x -\/x 2 -+- y 2 

y y 


ence dx = -f 


x dx -f y dy 

'- — — > 

V# 2 + y 2 


and dx = 


xdx-yydy 

V# 2 -+- y 2 


Integrating (2), 




and 


x = 
-c + 




z = + 


-yjx 2 + y 2 + c, 


— V# 2 + y 2 + c. 


Therefore (x - 


- c — s/x 2 + y 


V* 2 + 2 / 2 ) = 0, 




f- 


= c 2 - 


■2 ex. 





(2) 



or 



PROBLEMS. 

dv 2 

1 . -~ = ax. Ans. (y — c) 2 = 4 ax 9 . 
dx 2 

2. ^-5^ + 6 = 0. Ans. (y-2x + c)(y-3x + c) = 0. 
dx 2 dx 

3. x*%£ + 3xy^ + 2y 2 ==0. Ans. (xy + c) (afy + c) = 0. 

4. (a 2 + l)^!=l. ^s. c 2 e 2 *-2ca^ = l. 



5. 



f^(fc + ' V V^ + 2/)- ^ (^ 2 - 2 ^ + c )[ eX (^ + 2/- 1 ) + c ]- 



234 DIFFERENTIAL AND INTEGRAL CALCULUS. 



Art. 144. Differential Equations of the Second Order. 



I. Equations involving x and — - 2 only. 

ax 

The equation, if possible, is put in the form 

in which X is a function of x only. 

Integrating (1), f- = X, + C v (2) 

ax 

Integrating (2), y == X 2 + G x x + C 2 . (3) 

In (2) and (3), X r and X 2 are functions of a? only, and C x and C 2 are 
arbitrary constants. 
For example, given 

Then 



dx 2 


-ax*. 


d?y = 
dx 


— CtJb LvJby 


<fy = 
dx 


5 


dy = 


= ^^ + C,dx, 
5 


y = 


■£+<*+* 



d 2 v 
II. Equations involving y and — ^ cmfa/. 

The equation, if possible, is put in the form 



dx 2 
in which Yis a function of y only. 



&2L-Y, (4) 



Multiplying (4) by 2% 



dx dx 2 dx 



whence 2^d(^) = 2 Ydy. 

dx \dXj 



DIFFERENTIAL EQUATIONS. 235 

Integrating (5), 



dx) 
Separating the variables and integrating, 



2 JTdy + C v 



y J2fTdy + G 1 



For example, given — \ = a 2 y. 
dx 2 



Then 223 = 2^9, 

,7_. ,7 2 "^ -7..' 



dx dx 2 dx 






dx 



dy 



Va 2 y 2 + d 

III. Equations not involving y directly. 

The equation will be of the form F fx, % &] = 0. 

\ dx dx 2 ) 

Assume ^ = 2 , then & = *. 

Making these substitutions, an equation of the first order between 
z and x is obtained. 



= 0. 



For example, given _| + _|^ 1+/-^ 

Assume -^ = z, 

dx 

then d2: = - ?* dx. 

V(l + 2 2 ) 8 "' 

Integrating, 

— ^= = -^+0 1 = ^^when0 1 = 4; 

Vl + z- or a- a 2 



236 DIFFERENTIAL AND INTEGRAL CALCULUS. 

whence z = 



_dy _ c 2 — x 2 



Therefore y- ~ x ^ dx 



dx Va 4 - (c 2 - x 2 ) 

r (c 2 ~c 

J vV - ( 



x 2 ) 2 
IV. Equations not involving x directly. 

Assume -^ == z, 

dx 

then o^y^d^^olzdy^^ 

dx 2 dx dydx dy 

Making these substitutions, the independent variable is changed 
from x to y, and an equation of the first order between z and y is 
obtained. 



For example, given — -- — a( -^ ) =y. 
dx 2 \dxj 


(6) 


Substituting ^ = z, and ^ \= z — , 
dx dx 2 dy 




dz o 
z — — cur = y. 
dy 


(7) 


Assume z 2 = 2 v, then zdz = dv, and substituting in (7), 





dv — 2avdy = y dy. (8) 

Equation (8) is a linear equation of the first order and degree and 
is therefore integrable. 

PROBLEMS. 

Ans. y = x 3 + C L x -f- C 2 . 



1. 


d * y -6x 


2. 


dx 2 


3. 


d 2 y_ 1 
d% ^/ay 


4. 


dx 2 dx 



Ans. x = — =log — 1 f- C 2 . 

-y/2n VCiV' + l + l 

. 4/-r2/ 1,40a 80,/}, 4 0A1 , „ 
.4ns. « = Va -(# 2 +— r) T[y + i ) +^- 

-4ws. y — C x log aj + C 2 . 



DIFFERENTIAL EQUATIONS. 237 

5. (1 — X s ) — \ — x-^- = 2. Ans. y = (arc sin x) 2 + (7, arc sin x -J- C 2 . 
v J dx 2 dx * v y 

e. way = i +m. An S . 2 jl = q£+ g + C, 

7. y^ + (^) 2 = I- ^- / = or* + fta + a 2 . 

aar \dxj 

8. ,(l-togy)g+(l + logy/gY = 0. 

^4ns. log y — 1 = 



Cia; + C 2 

9 * y S ~ (t;) 2= ^ 2 log y " As * log y = ° ieX + ° 2e ~ x ' 

10. The equation of motion of a particle ascending in the air against 
the action of gravity is 

d 2 x _ _ _ jofdx^ 2 
d^~ 9 9 \dt 

Find the equation for the space described by the particle in terms of 

d,x 
the time; determining the constants of integration by making— =v 

when t = 0, and x = when t = 0. 

Ans. x = — - log (vh sin kgt + cos %£). 



APPENDIX. 



>XX< 



TABLE OF INTEGRALS. 



This table contains the principal integrals given in this book, to- 
gether with a few additional ones. The arrangement only partially 
follows the order in which the forms occur in the book, as convenience 
of reference is the first consideration. 

I. 

ELEMENTARY FORMS. 
1 . I (d u ■+■ dv — div) — u -f- v — w. 

2. J adx= ax. 4. I a — = a log x. 

3. Cadf(x) = af(x). 5. Cax n dx = ^j. 

EXPONENTIAL FORMS. 



6. ( a x log adx — a x . 7. I e*<:?a; = e x . 



TRIGONOMETRIC FORMS. 



8. r cosa?cto= sin a?. 10. J sin a; da; = — cos x. 

9. | sec 2 a; da; — tana;. 11. J cosec 2 a;da; = — cot a;. 



238 



APPENDIX. 239 

12. I secxt&nxdx = sec#. 14. I sin a; dx — vers #. 

13. I cosec x cot x dx = — cosec #. 15. I cos jc eto = — covers x. 

INVERSE TRIGONOMETRIC FUNCTIONS. 

C dx . n ~ C dx 

16. I — - ====t = arc sin x. 20. I = ^ == = arc sec x. 

J Vl — as 2 ^ a;Vic 2 — 1 

17. f ^ = arc cos a. 21. | -^ = arc cosec x. 

J a/1 _ ^ J x-yjtf - 1 



18. f_*L- = arotans. 22 - f , X = arc vers a;. 

Jl+a; 2 ^ V2 a; — a; 2 

19. f ^_ = arc cot x. 23. I = arc covers a;. 

J 1 + x 2 J 



dx 

V2¥ 



II. 

RATIONAL ALGEBRAIC FORMS. 
Expressions containing (a + foe). 

24. f-J2- = hog(a + bx). 
J a + bx b 

25. r_^--==I[a + 6a;-alog(a + 6aj)]. 

J a + bx b 2 

26. f\ /^_±£ dx = V(a + a*) (6 + a?) + (a — 6) log (Va + x + V& + x). 
J *b +x 

27 . P\F-=^ (to = V(a- »)(& + ») + (a + 6) arc sin X fe-±A 
J \& + a; Va + 6 

28. Cj^_ = J_ [i ( a + &aj)» _ 2 a (a + 6a) + a 2 log (a -f 6a?)]. 

nn C dx It a + bx 

29. I— = log — L 

J as(a-f&as) a a? 

' dx = _ JL , A 
a^(a + 6a?) ax a 2 



3 °- (v%-=--+ A >s a+te 

«/ x\a + 



240 DIFFERENTIAL AND INTEGRAL CALCULUS. 

31. f dx = L 

J (a -f 6a?) 2 b(a-\- 

32 - S 



(a + bx) 2 6 (a 4- bx) 
xdx 1 



(a + bx) 2 I/ 1 
2 clx _ 1 r 



log (a + bx) +_^/|. 
a + 6a? J 



33 ' f / ^TT ^ = h l a + to - 2 a l °8 (« + &*) ^r-1- 

J (a + 6a?) 2 6 3 1_ a + 6a?J 

34. C 

J a 



1 -, /a + bx 



a? (a + 6a?) 2 a (a + 6a?) a : 

Expressions containing (a + 6a? 2 ). 

35 . I — = - arc tan — 

J a 1 4- x l a a 

36. f^*L_= l]og« + *. 

*/ a J — ar J a a — x 

37. (*-*" 
%/ a 



;■= — — arc tan a; \/-, when a>0, and 6>0. 

^ Va6 Xa 



38 



' J a 2 



dx 



1 

= log 



6 2 a?- 2 a6 *" V a — 6a? 



a + 6a;' 



when a > 0, and 6 > 0. 



r adx -, x — a 

39. J- =log^— 

J x" — ar * a? + a 

„_ /* a?c£a? 1 , / 2 , c 



4- 6a? 2 26 

/• a? 2 aa? a? /a . lb 

41. I ,= \/— arc tana? \/— 

J a 4- 6a? 2 b \b 3 \o 

^o r ^ i r& , ' lb 

42. I — == — \l— arctanaj\/— 

J a? 2 (a + 6a? 2 ) aa? * a? * a 

43. f__^ = x + _J__ arc tan a? \P 

J fa 4- 6a? 2 ) 2 2a(a + bx 2 ) v^/Trf* ^ a 



(a 4- 6a? 2 ) 2 2 a (a 4- bx 2 ) 2 Va 3 6 

44 C dx _ 1 a? 2m -1 f_t^_ 

J (a 4- 6a? 2 ) w+1 "2ma(a+ 6a? 2 ) w 2 ma J (a 4- 6a? 2 



45. 



J (a 



x 2 dx 



(a 4- 6a? 2 ) ra+1 2 m6 (a 4- 6a? 2 )' 



+ 



1 f eta 

2 ra6 J (a -f 6. 



(a 4- 6a? 2 ) w 



APPENDIX. 241 



Expressions involving (a + bx n ). 

46. J x m (a + bx n ) p dx 

x m ~ n+1 (a + bx n ) p+1 — (m — n + 1)« Cx m - n (a + 6a n ) p cfo 
6 (?ip + m + 1) 

47. far ra (a + &a; n ) p cfce 

ar m+1 (a + &» n ) p+1 + b (m - rap — ?i — 1) fa;- m+n (a + bx n ) p dx 
— a (m — 1) 

48. Cx m (a + bx n ) p dx 

x m+1 (a + &£ n ) p + emp ( x m (a + bx^'^dx 
np + m + 1 

49. \ x m (a + bx n )~ p dx 

x m+1 (a + 6a n )- p+1 — (m + ?i + 1 — np) pc m (a + bx n )~ p+1 dx 
an (p — 1) 

III. 
IRRATIONAL ALGEBRAIC FUNCTIONS. 



Expressions containing Va -t- bx. 

50. ( y/a + bxdx = — V(a + 6x) 3 . 

51. I , =--Va + bx. 
J Va + bx & 

^ 15 b 2 

53. r * d * = _ 2 ( 2 «- to ) y^+to. 

JVa + te 36 

J x Wa + bx+VaJ 



242 



/: 



DIFFERENTIAL AND INTEGRAL CALCULUS. 

dx 1 , fya -{-bx — Va 



xVa + bx Va \Va + 6x + Va 
156 3 



— = — — log ( v ™ "^ ^ ^ ], when a > 0. 

Va + bx 



56. J ^ = 2(8a*-4^ + 3fry) V(n7 ^ 



57 . f g = 

J x 2 Va 4- 5a? 



Va 4- &£ & 



a# 



log 



Va 4- foe — Va 



2 Va 3 VVa + bx 



-VaV 

+ Va/ 



log (<e + V# 2 4- a 2 ). 



Expressions containing Va 2 4-a; 2 . 
58. fV^M 2 cto = | Va^T^ 2 " + % log (as + VoM^ 2 ). 

»/ Z Li 

6 o. r xdx =y^+ 

*^ Va 2 4-a; 2 

/~ Va 2 -f a; 2 -, /-«-: — o i /^a 4- Va 2 4- aA 
61. I — ■ — dx = Va 2 + x 2 — a log ( ^ — — 1 

/: 
/: 

■/: 



62 



63 



?£ 1, x 

^^rrr = - log ■ 

x Va 2 4- x 2 a a 4- Va 2 + a; 2 

***" = | Va 2 + ^ 2 - f log (a; + V^T^)- 

Va 2 + x 2 2 2 

;?£ Va 2 + x 2 



x 2 Va 2 + ar 2 ax 

65. fx 2 Va 2 + x 2 dx = -(2x 2 + a 2 )Vx 2 + a 2 - - log (x 4- Va 2 4- x 2 ). 
+j 8 8 

66 *" 



. f- 

67. C(a 2 + x 2 )*' to = | (2 » 2 + 5 a 2 ) Vi?+^ + ^ log (x + Va 2 + x 2 ). 
«/ 8 8 

68. f- 



Va 2 + x 2 , 1 



arVa 2 4- 



2aV 



^2 a 3 & V x y 



APPENDIX. 



243 



69 



/ dx 



V(a 2 + a? 8 ) 3 a 2 Va 2 + x 2 



70. I— — ^ = — — — — hlogicH-Va 2 + ^. 

c/ 3/ fly 



Expressions containing Va 2 — a£ 

71. I Va 2 — x 2 dx = %(x -Va 2 — af + a? arc sin - Y 

72 . f — ^ = arc sin -• 

J Va 2 - or 2 a 

73. Cx Va 2 - x 2 dx = - * V(a 2 - a 2 ) 3 . 

74. f- d * 1 



log 
a; Va 2 — ic 2 a Va + Va 2 — x 2 , 



J X X 

/x 



xdx 



— Va 2 — a? 2 . 



7 7 . Cx 2 Va 2 - ar 9 da; = - - V(a 2 - x 2 f + | 2 fa; Va 2 - x 2 + a 2 arc sin H ■ 



78 



79. 



Va 2 — x 2 



•C / 2 2 i 

= — -Va"- or + ■ 



arc sm 



f__^_ Va 2 

*^ or v'a 2 — x 2 a ~ 



/~ Va 2 — x 2 , Va 2 — x 2 • x 

80. I — dx = — arcsm-- 

J x 2 x a 

81. CV(a 2 - x 2 ) 3 dx = i ["a? V(a 2 - or 2 ) 3 + ^ Va^^ + 5|! 

82. f dg = * . 

^ V(a 2 -a 2 ) 3 aVa 2 -^ 



arcsm 



244 DIFFERENTIAL AND INTEGRAL CALCULUS. 

C x 2 dx x . x 

83. I — = — — arc sin- • 

J V(a 2 -x 2 ) 3 Va 2 -x 2 a 

g4 _ r x m dx = _x^ (a2 _ ^ i + m-1 a2 T m _ 2 ,, _ ^4 ^ 
J Va 2 — a 2 w- m J 



Expressions containing Va; 2 — a 2 . 

85 . I Vx 2 — a 2 dx = \ [x V# 2 — a 2 — a 2 log (x + V# 2 — a 2 )]. 

86. f ^ = log (a; + VaF=a*). 
J Vx 2 — a 2 

C XClx /-o q 

87. I — = V # — a . 
•^ Vic 2 — a 2 

88. j x V# 2 — a 2 dx = i V(x 2 — a 2 ) 3 . 

89 . r V^ 2 - a 2 ^ = y^ _ a 2 _ a arc cos a 
*/ sc a; 



90. I — = -arc sec-- 

*^ x 



dx 1 x 

z=== = - arc sec — 

Vx 2 - a 2 a a 



91. r aftfo = xy x2 _ a2 + a 2 log ^ + Va;2 _ a2v 
^ Vic 2 — a 2 * % 

dx Vx 2 — a? 



92. 



/: 



x 2 Vx 2 - a 2 a® 

93. CxWx 2 -a 2 dx = -(2x 2 - a 2 )Vx 2 -a 2 - --log (sj + VaT^a 2 ), 
%j 8 8 



/dx Vx 2 — a 2 1 a; 

- = v ^ „ „ h z—„ arc sec 



x 3 Vx 2 _ a 2 2 a¥ 2 a 3 
95. JV(x 2 - a 2 fdx 

= - |"xV(f - a 2 ) 3 - ^^ Vx 2 ^ 2 + ^ log (a + V^^c^) 



96 



X 



APPENDIX. 245 

dx x 



V(x* — a 2 ) 3 a 2 Vx 2 — a 2 

x 2 dx 

V(x 2 — a?y 



n „ C x 2 dx x , 

97 - J v^^? = -v^^ +los(a;+v ^^- 



98. *» 



'•/; 



(x 2 - a 2 y aV« 2 - a 2 

99. f (x 2 - a 2 ) 1 ^ = | (2a 2 - 5 a 2 )Vx 2 -a 2 + ^log (x+V^-a 2 ). 
«/ 8 8 



Expressions containing V2 ax ± x 2 . 

V2 ax — x 2 dx — V 2 ax — x 2 -\ arc ver — 

2 2 a 



101. i — : = = arc ver 



dx x 

V2 ax — x 2 



102. f g(to = 

^ V2 ax - a 2 



— — V2 ax — x 2 + ct arc ver — ■ 

a 



103. 



£ 



j^ _ V2 ax - x 2 



:V2 ax — x 2 ax 

,_„ r /o o^r 3a 2 + ax — 2x 2 /~ -s , a 2 x 

104. I xv z ax — x^ax = V2 ax — x l -\ — arc ver — 

J 6 2 a 

•, rtl - C^2ax — x 2 -, /o -a . x 

105. I — dx = V2ax — x 2 ^- aarc ver — 

J x a 

107. r *» — = g - a . 

J (2ax- a?)* oV2 ax - a 2 



108. f ■"*■ 



(2 ax — x 2 )* a^/2 ax — x 2 



. f— = ^ = =log(x + a-fV2ax + x 2 ). 



109 

V2 ax + x 2 



246 DIFFERENTIAL AND INTEGRAL CALCULUS. 

x 2 dx x + 3 a /T) 



110 



111 






v2 ax — x 2 

xhlx 
V2 ax — x 2 



2 



a/2 ax — x 2 + % a 2 arc ver -. 



' + | ax + 1 a 2 )V2ax — x 2 + | a 3 arc ver _. 



Expressions containing Va + 5x ± 



ex . 



112 



113 



•/ 



da; 



Va 4- 6x 4- ex 2 Vc 



= — log (2 ex + b + 2 Vc Va + bx + ex 2 ). 



/Va 4- bx + cx 2 dx = "*~ Va 4- &x 4- ex 2 



4c 

6 2 — 4 ac 



log (2 ex + 6 + 2 Vc Va + bx + ex 2 ). 



114 



115 



■/: 



dx 



1 . 2cx-6 
= — arc sin 



116 



./vs 



/ 



Va + 6x — ex 2 Vc Vfr 2 4- 4ac 

2cx-6 



-{-bx — cx 2 dx = 



4c 

o 2 4- 4 ac 



Va 4- 6x — ex 2 
2 ex — & 



arc sin 



V& 2 + 4ac 



xdx 



Va + 6x + ex 2 



Va 4- 6x 4- ex 2 



b_ 

2c 



117 



' J v^ 



x 2 dx 



— = log (2 ex 4- 6 4-2 Vc Va + 6x + ex 2 ) 
Vc 



o/x 3b 



+ 



Va -|- &x 4- ex 2 1 
3 & 2 a\ T dx 



+ bx+cx 2 V2c 4c 2 



c 2 2 c 



r 

J Va~+ 



6x 4- ex 2 



118, 



*^ Va 



x n dx 



x" _1 Va + 5x + cx 2 



4- &x 4- ex 2 
n — 1 a 



•Jf: 



WC 

x n " 2 dx 



Va 4- bx 4- ex' 



2n cJ 



x n ~ J dx 



Va 4- 6x -f ex 2 



APPENDIX. 



247 



IV. 



TRIGONOMETRIC AND TRANSCENDENTAL FUNCTIONS. 



119. | sm 2 xdx = ^x — \ sin2a;. 



120. J cos 2 xdx = -J-a? + \ sin2». 

121. J tan x dx = log sec x. 

122. I cotxdx = log sin a?. 

123. f-4^ = log tan i a;. 
J sin a; 

124. CJ*~ =logtaaf^ + 4.A 
•/ cos a; \4 / 

125 . I cosec x dx = log tan A- a;. 

126. f- 
»/ a 



+ & cos Va 2 - 6 2 



arc tan 



a — 6\ 3 
tan- 



when a > b, 



Vfr + a + Vfr — a tan 



log 



127 



V6 2 - a 2 
. I x sin xdx= sin a; — a; cos a;. 

128. I x 2 smxdx = 2#sina; — (x 2 — 2)cossc. 

129. I x cos x dx = cos jc + x sin a;. 

130. I ar 2 cos a; da; = 2 a? cos a; -h (x 2 — 2) sin x. 

_ - /^sinaj 7 a; 3 . a; 5 a; 7 

131. I dx=x h •••• 

J x 3[3 5[5 7 [7 



'., when a<b. 



V6 + a — V 6 — a tan x 



248 DIFFERENTIAL AND INTEGRAL CALCULUS. 

-. /^< to =io S ,.-4 + 4-4 + .... 

133. | arc sin x dx = x arc sin a? + VI — x 2 . 

134. I arccosxda? = x arc cos a? — Vl — x 2 . 

135. I arctanxcte = #arctana? — ilog(l -f x 2 ). 

136. I arc cotan x dx = x arc cot x -f-ilog(l + x 2 ). 

137. f arcversicdx = (x — 1) arc vers x + V2sc— a? 2 . 

138. I log a? c£x = x log a; — x. 

/' dx 1 1 
= log (log X) + log X + -„ log 2 <B + 7— -rs log 3 X + • ■ 
log <B 2^ 2 • <T 

140. I = log (log 05). 

J x log a? 

xe ax dx = ^— (a x — 1). 

a 2 v 

-. >. o C ax • i e ax (a sin a? — cos x) 

142. I e ax sm £ a# = — ^ ^ 

J a 2 + l 

- ,. „ r«, 7 6 ax (a cos a; + sin x) 

143 . I e ax cos x dx = — ^ — -± L. 

J a 2 + l 

144. Ce a *logxdx = eaXlogx -* f— da. 

145. f a m e"cfo = =^ - - ftf-Vdas. 
*/ a a J 

146. (V log xdx = x m+1 \^^- - - — i— - |. 
J |_m + 1 (m + 1) 2 J 

-.j~ C ' « ? sin n_1 a? cos a? . n — 1 /" ■ n _ 2 7 

147. I sin" xdx = \- - I sin" J x dx. 

J n 7i J 



APPENDIX. 249 

148. I cos n x clx = - cos n_1 x sin x -\ | cos n ~ 2 xclx. 

J n n J 

, * n C m - n 7 cos m_1 x sin n+1 x . m — 1 C m 2 • « 7 

149. I cos m x sin n x clx = 1 I cos" 1- - x sin" x clx 

J m + n m + nJ 

sin n_1 x cos m+1 x . n — 1 C m • « 2 7 

= 1 I cos' n x sm n ~ i! xclx. 

m + n m -{- nJ 

150. I tan n x clx = - — I tan n ~ 2 x clx. 

J n — 1 J 

x m loer n x clx = — log' 1 x I x m log" -1 x clx. 

m. + l m + lJ 

Cx m clx _ aT+^ m + 1 r a^cta 

J log n £ (?i — 1) log n_1 x n — 1 J log n_1 X 

a mx x n clx = -^— - — ( a mx x n ~ x clx. 

mloga m log a J 

Ca x clx cf log a Ca x clx 

J x m (m — l)x m ~ 1 m — lJ a;" 1-1 

e ax cos n_1 x (ft cos x -f- n sin x 



/• 



155. I e ax cos n x clx = 



ft 2 -f n- 



~-±L fe ax cos' 1 - 2 xclx. 
ft 2 -\- n 2 J 



n (n — 1) 
ft 2 + n 2 

/ x m-l 
x m cos ft.s cfa = — - (ax sin ax -\- m cos «x) 
ft 2 

m (m — 1) C m-2 7 

^ '- I X m * COS OJC ft£. 

ft 2 J 



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